Eureka Math

Engage NY Eureka Math 8th Grade Module 2 Lesson 3 Answer Key

Eureka Math Grade 8 Module 2 Lesson 3 Exercise Answer Key

Exercise 1.
Draw a line passing through point P that is parallel to line L. Draw a second line passing through point P that is parallel to line L and that is distinct (i.e., different) from the first one. What do you notice?

Answer:

Students should realize that they can only draw one line through point P that is parallel to L.

Exercises 2–4 (9 minutes)
Students complete Exercises 2–4 independently in preparation for the discussion that follows.

Exercise 2.
Translate line L along the vector \(\overrightarrow{A B}\). What do you notice about L and its image, L’?

Answer:

L and L’ coincide. L=L’.

Exercise 3.
Line L is parallel to vector \(\overrightarrow{A B}\). Translate line L along vector (AB) ⃗. What do you notice about L and its image, L’?

Answer:

L and L’ coincide, again. L=L’.

Exercise 4.
Translate line L along the vector \(\overrightarrow{A B}\). What do you notice about L and its image, L’?

Answer:

L || L’

Exercises 5–6 (5 minutes)
Students complete Exercises 5 and 6 in pairs or small groups.

Exercise 5.
Line L has been translated along vector \(\overrightarrow{A B}\), resulting in L’. What do you know about lines L and L’?

Answer:
L || T(L)

Question 6.
Translate L₁ and L₂ along vector \(\overrightarrow{D E}\). Label the images of the lines. If lines L₁ and L₂ are parallel, what do you know about their translated images?

Answer:

Since L₁ || L₁, then (L₁)’ || (L₂)’.

Eureka Math Grade 8 Module 2 Lesson 3 Exit Ticket Answer Key

Question 1.
Translate point Z along vector \(\overrightarrow{A B}\). What do you know about the line containing vector \(\overrightarrow{A B}\) and the line formed when you connect Z to its image Z’?

Answer:
The line containing vector \(\overrightarrow{A B}\) and ZZ’ is parallel.

Question 2.
Using the above diagram, what do you know about the lengths of segment ZZ’ and segment AB?
Answer:
The lengths are equal: |ZZ’|=|AB|.

Question 3.
Let points A and B be on line L and the vector \(\overrightarrow{A C}\) be given, as shown below. Translate line L along vector \(\overrightarrow{A C}\). What do you know about line L and its image, L’? How many other lines can you draw through point C that have the same relationship as L and L’? How do you know?

Answer:

L and L’ are parallel. There is only one line parallel to line L that goes through point C. The fact that there is only one line through a point parallel to a given line guarantees it.

Eureka Math Grade 8 Module 2 Lesson 3 Problem Set Answer Key

Question 1.
Translate ∠XYZ, point A, point B, and rectangle HIJK along vector \(\overrightarrow{E F}\). Sketch the images, and label all points using prime notation.

Answer:

Question 2.
What is the measure of the translated image of ∠XYZ? How do you know?
Answer:
The measure is 38°. Translations preserve angle measure.

Question 3.
Connect B to B’. What do you know about the line that contains the segment formed by BB’ and the line containing the vector \(\overrightarrow{E F}\)?
Answer:
\(\overleftrightarrow{\boldsymbol{B B}^{\prime}} \| \overleftrightarrow{\boldsymbol{E F}}\)

Question 4.
Connect A to A’. What do you know about the line that contains the segment formed by AA’ and the line containing the vector \(\overrightarrow{\boldsymbol{E F}}\)?
Answer:
\(\overleftrightarrow{A A^{\prime}}\) and \(\overleftrightarrow{\boldsymbol{E F}}\) coincide.

Question 5.
Given that figure HIJK is a rectangle, what do you know about lines that contain segments HI and JK and their translated images? Explain.
Answer:
Since HIJK is a rectangle, I know that \(\overleftrightarrow{\boldsymbol{H} \boldsymbol{I}}\) || \(\overleftrightarrow{\boldsymbol{J K}}\). Since translations map parallel lines to parallel lines, then \(\overleftrightarrow{\boldsymbol{H}^{\prime} \boldsymbol{I}^{\prime}}\)||\(\overleftrightarrow{\boldsymbol{J}^{\prime} \boldsymbol{K}^{\prime}}\).

Engage NY Eureka Math 8th Grade Module 2 Lesson 2 Answer Key

Eureka Math Grade 8 Module 2 Lesson 2 Exercise Answer Key

Exercise 1.
Draw at least three different vectors, and show what a translation of the plane along each vector looks like. Describe what happens to the following figures under each translation using appropriate vocabulary and notation as needed.

Answer:
Answers will vary.

Exercise 2.
The diagram below shows figures and their images under a translation along \(\overrightarrow{H I}\). Use the original figures and the translated images to fill in missing labels for points and measures.

Answer:
Solutions are in red, below.

Eureka Math Grade 8 Module 2 Lesson 2 Exit Ticket Answer Key

Question 1.
Name the vector in the picture below.

Answer:
\(\overrightarrow{Q P}\)

Question 2.
Name the vector along which a translation of a plane would map point A to its image T(A).

Answer:
\(\overrightarrow{S R}\)

Question 3.
Is Maria correct when she says that there is a translation along a vector that maps segment AB to segment CD? If so, draw the vector. If not, explain why not.

Answer:

Yes. Accept any vector that would translate the segment AB to segment CD. A possible vector is shown in red, above.

Question 4.
Assume there is a translation that maps segment AB to segment CD shown above. If the length of segment CD is 8 units, what is the length of segment AB? How do you know?
Answer:
The length of CD must be 8 units in length because translations preserve the lengths of segments.

Eureka Math Grade 8 Module 2 Lesson 2 Problem Set Answer Key

Question 1.
Translate the plane containing Figure A along \(\overrightarrow{A B}\). Use your transparency to sketch the image of Figure A by this translation. Mark points on Figure A, and label the image of Figure A accordingly.

Answer:
Marked points will vary. Verify that students have labeled their points and images appropriately.

Question 2.
Translate the plane containing Figure B along \(\overrightarrow{B A}\). Use your transparency to sketch the image of Figure B by this translation. Mark points on Figure B, and label the image of Figure B accordingly.

Answer:
Marked points will vary. Verify that students have labeled their points and images appropriately.

Question 3.
Draw an acute angle (your choice of degree), a segment with length 3 cm, a point, a circle with radius 1 in., and a vector (your choice of length, i.e., starting point and ending point). Label points and measures (measurements do not need to be precise, but your figure must be labeled correctly). Use your transparency to translate all of the figures you have drawn along the vector. Sketch the images of the translated figures and label them.
Answer:
Drawings will vary. Note: Drawing is not to scale.

Question 4.
What is the length of the translated segment? How does this length compare to the length of the original segment? Explain.
Answer:
The length is 3 cm. The length is the same as the original because translations preserve the lengths of segments.

Question 5.
What is the length of the radius in the translated circle? How does this radius length compare to the radius of the original circle? Explain.
Answer:
The length is 1 in. The length is the same as the original because translations preserve lengths of segments.

Question 6.
What is the degree of the translated angle? How does this degree compare to the degree of the original angle? Explain.
Answer:
Answers will vary based on the original size of the angle drawn. The angles will have the same measure because translations preserve degrees of angles.

Question 7.
Translate point D along vector \(\overrightarrow{A B}\), and label the image D’. What do you notice about the line containing vector \(\overrightarrow{A B}\) and the line containing points D and D’? (Hint: Will the lines ever intersect?)

Answer:

The lines will be parallel.

Question 8.
Translate point E along vector \(\overrightarrow{A B}\), and label the image E’. What do you notice about the line containing vector \(\overrightarrow{A B}\) and the line containing points E and E’?

Answer:

The lines will coincide.

Engage NY Eureka Math 8th Grade Module 2 Lesson 1 Answer Key

Eureka Math Grade 8 Module 2 Lesson 1 Exploratory Challenge Answer Key

a. Describe, intuitively, what kind of transformation is required to move the figure on the left to each of the figures (1)–(3) on the right. To help with this exercise, use a transparency to copy the figure on the left. Note: Begin by moving the left figure to each of the locations in (1), (2), and (3).

Answer:
Slide the original figure to the image (1) until they coincide. Slide the original figure to (2), and then flip it so they coincide. Slide the original figure to (3), and then turn it until they coincide.

b. Given two segments AB and CD, which could be very far apart, how can we find out if they have the same length without measuring them individually? Do you think they have the same length? How do you check? In other words, why do you think we need to move things around on the plane?

Answer:
We can trace one of the segments on the transparency and slide it to see if it coincides with the other segment. We move things around in the plane to see if they are exactly the same. This way, we don’t have to do any measuring.

Eureka Math Grade 8 Module 2 Lesson 1 Problem Set Answer Key

Question 1.
Using as much of the new vocabulary as you can, try to describe what you see in the diagram below.

Answer:
There was a transformation, F, that moved point A to its image F(A) and point B to its image F(B). Since a transformation preserves distance, the distance between points A and B is the same as the distance between points F(A) and F(B).

Question 2.
Describe, intuitively, what kind of transformation is required to move Figure A on the left to its image on the right.

Answer:
First, I have to slide Figure A so that the point containing two dots maps onto the Image of A in the same location; next, I have to turn (rotate) it so that Figure A maps onto Image of A; finally, I have to flip the figure over so the part of the star with the single dot maps onto the image.

Eureka Math Grade 8 Module 2 Lesson 1 Exit Ticket Answer Key

First, draw a simple figure and name it Figure W. Next, draw its image under some transformation (i.e., trace your
Figure W on the transparency), and then move it. Finally, draw its image somewhere else on the paper.
Describe, intuitively, how you moved the figure. Use complete sentences.
Answer:
Accept any figure and transformation that is correct. Check for the same size and shape. Students should describe the movement of the figure as sliding to the left or right, turning to the left or right, or flipping, similar to how they described the movement of figures in the exercises of the lesson.

Engage NY Eureka Math 8th Grade Module 1 End of Module Assessment Answer Key

Eureka Math Grade 8 Module 1 End of Module Assessment Task Answer Key

Question 1.
You have been hired by a company to write a report on Internet companies’ Wi-Fi ranges. They have requested that all values be reported in feet using scientific notation.

Ivan’s Internet Company boasts that its wireless access points have the greatest range. The company claims that you can access its signal up to 2,640 feet from its device. A competing company, Winnie’s Wi-Fi, has devices that extend to up to 2\(\frac{1}{2}\) miles.

a. Rewrite the range of each company’s wireless access devices in feet using scientific notation, and state which company actually has the greater range (5,280 feet =1 mile).
Answer:
Ivan’s Range: 2,640 = 2.64 × 10³ ft.
Winner’s Range: (2.5)5280 = 13200 = 1.32 × 10⁴ ft.
Winner’s wi-fi has the greater range.

b. You can determine how many times greater the range of one Internet company is than the other by writing their ranges as a ratio. Write and find the value of the ratio that compares the range of Winnie’s wireless access devices to the range of Ivan’s wireless access devices. Write a complete sentence describing how many times greater Winnie’s Wi-Fi range is than Ivan’s Wi-Fi range.
Answer:
winnie to ivan’s ratio – (1.32 × 10⁴) : (2.64 × 10³)
Value of ratio – \(\frac{1.32 \times 10^{4}}{2.64 \times 10^{3}}\) = \(\frac{1.32}{2.64}\) × \(\frac{10^{4}}{10^{3}}\) = \(\frac{1}{2}\) × 10 = 5
winnie’s wi-fi is 5 times greater in range than ivan’s internet company.

c. UC Berkeley uses Wi-Fi over Long Distances (WiLD) to create long-distance, point-to-point links. UC Berkeley claims that connections can be made up to 10 miles away from its device. Write and find the value of the ratio that compares the range of Ivan’s wireless access devices to the range of Berkeley’s WiLD devices. Write your answer in a complete sentence.
Answer:
(10)5280 = 52800 = 5.28 × 10⁴
Ivan’s to berkeley ratio: (2.64 × 10³): (5.28 × 10⁴)
Value of ratio- \(\frac{2.64 \times 10^{3}}{5.28 \times 10^{4}}\) = \(\frac{2.64}{5.28}\)×\(\frac{10^{3}}{10^{4}}\)= \(\frac{1}{2}\)×\(\frac{1}{10}\)= \(\frac{1}{20}\)
Ivan’s internet devices have a range \(\frac{1}{20}\) the range of UC Berkeley wild devices.

Question 2.
There is still controversy about whether or not Pluto should be considered a planet. Although planets are mainly defined by their orbital path (the condition that prevented Pluto from remaining a planet), the issue of size is something to consider. The table below lists the planets, including Pluto, and their approximate diameters in meters.

a. Name the planets (including Pluto) in order from smallest to largest.
Answer:
Pluto, mercury, mars, venus, earth, neptune, uranus, saturn, jupiter.

b. Comparing only diameters, about how many times larger is Jupiter than Pluto?
Answer:
\(\frac{1.43 \times 10^{8}}{2.3 \times 10^{6}}\) = \(\frac{1.43}{2.3}\) × \(\frac{10^{8}}{10^{6}}\)
≈ 0.622 × 10²
≈ 62.2
The diameter of jupiter is about 62 times larger than pluto.

c. Again, comparing only diameters, find out about how many times larger Jupiter is compared to Mercury.
Answer:
\(\frac{1.43 \times 10^{8}}{4.88 \times 10^{6}}\) = \(\frac{1.43}{4.88}\) × \(\frac{10^{8}}{10^{6}}\)
≈ 0.293 × 10²
≈ 29.3
The diameter of jupiter is about 29 times larger than mercury.

d. Assume you are a voting member of the International Astronomical Union (IAU) and the classification of Pluto is based entirely on the length of the diameter. Would you vote to keep Pluto a planet or reclassify it? Why or why not?
Answer:
I would vote to reclassify it. Knowing that jupiter is 29 times larger than mercury means mercury is pretty small. Jupiter is 42 times larger than pluto, which means pluto is even. Smaller than mercury. For that reason vote that the length of the diameter of pluto is too small compared to other planets (Even the small one).

e. Just for fun, Scott wondered how big a planet would be if its diameter was the square of Pluto’s diameter. If the diameter of Pluto in terms of meters were squared, what would the diameter of the new planet be? (Write answer in scientific notation.) Do you think it would meet any size requirement to remain a planet? Would it be larger or smaller than Jupiter?
Answer:
(2.3 × 10⁶)² = 2.3² × (10⁶)²
= 5.29 × 10¹²
Yes, 5.29 × 10¹² would likely meet any size requirement for planets. It would be larger than jupiter.

Question 3.
Your friend Pat bought a fish tank that has a volume of 175 liters. The brochure for Pat’s tank lists a “fun fact” that it would take 7.43×10¹⁸ tanks of that size to fill all the oceans in the world. Pat thinks the both of you can quickly calculate the volume of all the oceans in the world using the fun fact and the size of her tank.

a. Given that 1 liter =1.0×10^-12 cubic kilometers, rewrite the size of the tank in cubic kilometers using scientific notation.
Answer:
175 litres = 175(1.0×10^-12) cubic kilometers
= 175 ×10^-12 km³
= 1.75 × 10^-10 km³

b. Determine the volume of all the oceans in the world in cubic kilometers using the “fun fact.”
Answer:
(1.75 × 10^-10)(7.43×10¹⁸) = (1.75× 7.43)(10^-10×10¹⁸)
= 13.0025×10⁸
= 1.30025×10⁹
The volume of all the oceans in the world is (1.30025×10⁹)km³.

c. You liked Pat’s fish so much you bought a fish tank of your own that holds an additional 75 liters. Pat asked you to figure out a different “fun fact” for your fish tank. Pat wants to know how many tanks of this new size would be needed to fill the Atlantic Ocean. The Atlantic Ocean has a volume of 323,600,000 cubic kilometers.
Answer:
Tank: 175+75 = 250 liters
250 liters = 250(1.0×10^-12) km³
= 2.50 ×10^-12
= 2.5 × 10^-10
Atlantic ocean: 323,600,000
= 3.236 × 10⁸ km³

\(\frac{3.236 \times 10^{8}}{2.5 \times 10^{-10}}\) = \(\frac{3.236}{2.5}\) × \(\frac{10^{8}}{10^{-10}}\)
= 1.2944×10¹⁸
It would take 1.2944×10¹⁸ tanks (of size 250 liters) to fill the atlantic ocean.

Engage NY Eureka Math 8th Grade Module 1 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 1 Mid Module Assessment Task Answer Key

Question 1.
The number of users of social media has increased significantly since the year 2001. In fact, the approximate number of users has tripled each year. It was reported that in 2005 there were 3 million users of social media.

a. Assuming that the number of users continues to triple each year, for the next three years, determine the number of users in 2006, 2007, and 2008.
Answer:
2006-9 MILLION
2007-27 MILLION
2008-81 MILLION

b. Assume the trend in the numbers of users tripling each year was true for all years from 2001 to 2009. Complete the table below using 2005 as year 1 with 3 million as the number of users that year.

Answer:

c. Given only the number of users in 2005 and the assumption that the number of users triples each year, how did you determine the number of users for years 2, 3, 4, and 5?
Answer:
I MULTIPLIED THE PRECEDING YEAR’S NUMBER OF USERS BY 3.

d. Given only the number of users in 2005 and the assumption that the number of users triples each year, how did you determine the number of users for years 0, -1, -2, and -3?
Answer:
I Divided The Next Year’s Number of users by 3.

e. Write an equation to represent the number of users in millions, N, for year t, t≥-3.
Answer:
N=3t

f. Using the context of the problem, explain whether or not the formula N=3^t would work for finding the number of users in millions in year t, for all t≤0.
Answer:
We only know that the number of users has tripled each year in the time frame of 2001 to 2009. For that reason, we cannot rely on the formula, N = 3t, to work for all t≤0, just to t=-3, which is the year 2001.

g. Assume the total number of users continues to triple each year after 2009. Determine the number of users in 2012. Given that the world population at the end of 2011 was approximately 7 billion, is this assumption reasonable? Explain your reasoning.
Answer:
2012 is t=8, so when t=8 in N=6,561,000,000. The number of users in 2012, 6,561,000,000. Does not exceed the world population of 7 billion, therefore it is possible to have that number of users. But 6,561,000,000 is Approximately 94%. of the world’s population. The number of users is likely less than that due to poverty, illness, infancy, etc. The Assumption is possible, but not reasonable.

Question 2.
Let m be a whole number.
a. Use the properties of exponents to write an equivalent expression that is a product of unique primes, each raised to an integer power.
\(\frac{6^{21} \cdot 10^{7}}{30^{7}}\)
Answer:
\(\frac{(3 \cdot 2)^{21} \cdot 10^{7}}{(3 \cdot 10)^{7}}\)=\(\frac{3^{21} \cdot 2^{21} \cdot 10^{7}}{3^{7} \cdot 10^{7}}\)
= 3^21-7.2²¹.10^7-7
= 3¹⁴.2²¹.10⁰
= 3¹⁴.2²¹

b. Use the properties of exponents to prove the following identity:
\(\frac{6^{3 m} \cdot 10^{m}}{30^{m}}\) = 2^3m.3^2m
Answer:
\(\frac{6^{3 m} \cdot 10^{m}}{30^{m}}\) = \(\frac{(3.2)^{3 m} \cdot 10^{m}}{(3 \cdot 10)^{m}}\)
= \(\frac{3^{3 m} \cdot 2^{3 m} \cdot 10^{m}}{3^{m} \cdot 10^{m}}\)
= 3^3m-3.2^3m.10^m-m
= 3^2m.2^3m
= 2^3m.3^2m

c. What value of m could be substituted into the identity in part (b) to find the answer to part (a)?
Answer:
2^3m.3^2m = 2²¹.3¹⁴
3m = 21
m = 7
2m = 14
m = 7
Therefore, m = 7

Question 3.
Jill writes 2³ ∙4³ =8⁶ and the teacher marked it wrong. Explain Jill’s error.
Answer:
Jill multiplied the bases, 2 and 4, and added the exponents. You can only add the exponents when the bases being multiplied are the same.

b. Find n so that the number sentence below is true:
2³ ∙4³ =2³ ∙2ⁿ =2⁹
Answer:
4³ = 4.4.4
= (2.2)(2.2)(2.2)
= 2⁶
Therefore
2³.4³ = 2³.2⁶ = 2⁹
so
n = 6

c. Use the definition of exponential notation to demonstrate why 2³ ∙4³ =2⁹ is true.
Answer:
4³=2⁶, 2³.4³ = 2⁹ is equivalent to 2³.2⁶ = 2⁹
By definition of exponential notation:

d. You write 7⁵ ∙7^-9 =7^-4 . Keisha challenges you, “Prove it!” Show directly why your answer is correct without referencing the laws of exponents for integers; in other words, x^a ∙x^b =x^a+b for positive numbers x and integers a and b.
Answer:
7⁵.7^-9 = 7⁵.\(\frac{1}{7^{9}}\) by definition
= \(\frac{7^{5}}{7^{9}}\) by product formula
= \(\frac{7^{5}}{7^{5.7^{4}}}\) by x^m.xⁿ = x^m+n for x>0, m,n≥0
= \(\frac{1}{7^{4}}\) by equivalent fractions
= 7^-4 by definition.

Engage NY Eureka Math 8th Grade Module 1 Lesson 13 Answer Key

Eureka Math Grade 8 Module 1 Lesson 13 Exercise Answer Key

Exercise 1.
The Fornax Dwarf galaxy is 4.6×10⁵ light-years away from Earth, while Andromeda I is 2.430×10⁶ light-years away from Earth. Which is closer to Earth?
Answer:
2.430×10⁶=2.430×10×10⁵=24.30×10⁵
Because 4.6<24.30, then 4.6×10⁵<24.30×10⁵, and since 24.30×10⁵=2.430×10⁶, we know that 4.6×10⁵<2.430×10⁶. Therefore, Fornax Dwarf is closer to Earth.

Exercise 2.
The average lifetime of the tau lepton is 2.906×10^-13 seconds, and the average lifetime of the neutral pion is 8.4×10^-17 seconds. Explain which subatomic particle has a longer average lifetime.
Answer:
2.906×10^-13=2.906×10⁴×10^-17=29,060×10^-17
Since 8.4<29,060, then 8.4×10^-17<29,060×10^-17, and since 29,060×10^-17=2.906×10^-13, we know that 8.4×10^-17< 2.906×10^-13. Therefore, tau lepton has a longer average lifetime.
This problem, as well as others, can be solved using an alternate method. Our goal is to make the magnitude of the numbers we are comparing the same, which will allow us to reduce the comparison to that of whole numbers.
Here is an alternate solution:
8.4×10^-17=8.4×10^-4×10^-13=0.000 84×10^-13.
Since 0.00084<2.906, then 0.000 84×10^-13<2.906×10^-13, and since 0.000 84×10^-13=8.4×10^-17, we know that 8.4×10^-17< 2.906×10^-13. Therefore, tau lepton has a longer average lifetime.

Exploratory Challenge 1/Exercise 3.
THEOREM: Given two positive numbers in scientific notation, a×10^m and b×10ⁿ, if m<n, then a×10^m<b×10ⁿ.

Prove the theorem.
If m<n, then there is a positive integer k so that n=k+m.
By the first law of exponents (10) in Lesson 5, b×10ⁿ=b×10^k×10^m=(b×10^k)×10^m. Because we are comparing with a×10^m, we know by (1) that we only need to prove a<(b×10^k). By the definition of scientific notation, a<10 and also (b×10^k)≥10 because k≥1 and b≥1, so that (b×10^k)≥1×10=10. This proves a<(b×10^k), and therefore, a×10^m<b×10ⁿ.
Explain to students that we know that a<10 because of the statement given that a×10^m is a number expressed in scientific notation. That is not enough information to convince students that a<b×10^k; therefore, we need to say something about the right side of the inequality. We know that k≥1 because k is a positive integer so that n=k+m. We also know that b≥1 because of the definition of scientific notation. That means that the minimum possible value of b×10^k is 10 because 1×10¹=10. Therefore, we can be certain that a<b×10^k.
Therefore, by (1), a×10^m<(b×10^k)×10^m. Since n=k+m, we can rewrite the right side of the inequality as
b×10ⁿ, and finally a×10^m<b×10ⁿ.

Exercise 4.
Compare 9.3×10²8 and 9.2879×10²8.
Answer:
We only need to compare 9.3 and 9.2879. 9.3×10⁴=93,000 and 9.2879×10⁴=92,879, so we see that
93,000>92,879. Therefore, 9.3×10²⁸>9.2879×10²⁸.

Exercise 5.
Chris said that 5.3×10⁴¹<5.301×10⁴1 because 5.3 has fewer digits than 5.301. Show that even though his answer is correct, his reasoning is flawed. Show him an example to illustrate that his reasoning would result in an incorrect answer. Explain.
Answer:
Chris is correct that 5.3×10⁴¹<5.301×10⁴¹, but that is because when we compare 5.3 and 5.301, we only need to compare 5.3×10³ and 5.301×10³ (by (1) above). But, 5.3×10³<5.301×10³ or rather 5,300<5,301, and this is the reason that 5.3×10⁴¹<5.301×10⁴¹. However, Chris’s reasoning would lead to an incorrect answer for a problem that compares 5.9×10⁴¹ and 5.199×10⁴¹. His reasoning would lead him to conclude that
5.9×10⁴¹<5.199×10⁴¹, but 5,900>5,199, which is equivalent to 5.9×10³>5.199×10³. By (1) again, 5.9>5.199, meaning that 5.9×10⁴¹>5.199×10⁴¹.

Exploratory Challenge 2/Exercise 6.
You have been asked to determine the exact number of Google searches that are made each year. The only information you are provided is that there are 35,939,938,877 searches performed each week. Assuming the exact same number of searches are performed each week for the 52 weeks in a year, how many total searches will have been performed in one year? Your calculator does not display enough digits to get the exact answer. Therefore, you must break down the problem into smaller parts. Remember, you cannot approximate an answer because you need to find an exact answer. Use the screen shots below to help you reach your answer.

Answer:
First, I need to rewrite the number of searches for each week using numbers that can be computed using my calculator.
35 939 938 877=35 939 000 000+938 877
=35 939×10⁶+938 877
Next, I need to multiply each term of the sum by 52, using the distributive law.
(35 939×10⁶+938 877)×52=(35 939×10⁶)×52+(938 877×52)
By repeated use of the commutative and associative properties, I can rewrite the problem as
(35 939×52)×10⁶+(938 877×52).
According to the screen shots, I get
1 868 828×10⁶+48 821 604=1 868 828 000 000+48 821 604
=1 868 876 821 604.
Therefore, 1,868,876,821,604 Google searches are performed each year.

Yahoo! is another popular search engine. Yahoo! receives requests for 1,792,671,335 searches each month. Assuming the same number of searches are performed each month, how many searches are performed on Yahoo! each year? Use the screen shots below to help determine the answer.

Answer:
First, I need to rewrite the number of searches for each month using numbers that can be computed using my calculator.
1 792 671 335=1 792 000 000+671 335
=1 792×10⁶+671 335.
Next, I need to multiply each term of the sum by 12, using the distributive law.
(1 792×10⁶+671 335)×12=(1 792×10⁶)×12+(671 335×12).
By repeated use of the commutative and associative properties, I can rewrite the problem as
(1 792×12)×10⁶+(671 335×12)
According to the screen shots, I get
21 504×10⁶+8 056 020=21 504 000 000+8 056 020
=21 512 056 020
Therefore, 21,512,056,020 Yahoo! searches are performed each year.

Eureka Math Grade 8 Module 1 Lesson 13 Exit Ticket Answer Key

Question 1.
Compare 2.01×10¹5 and 2.8×10¹3. Which number is larger?
Answer:
2.01×10¹⁵=2.01×10²×10¹³=201×10¹³
Since 201>2.8, we have 201×10¹³>2.8×10¹³, and since 201×10¹³=2.01×10¹⁵, we conclude 2.01×10¹⁵>2.8×10¹³.

Question 2.
The wavelength of the color red is about 6.5×10^^-9 m. The wavelength of the color blue is about 4.75×10^-9 m. Show that the wavelength of red is longer than the wavelength of blue.
Answer:
We only need to compare 6.5 and 4.75:
6.5×10^-9=650×10^-7 and 4.75×10^-9=475×10^-7, so we see that 650>475.
Therefore, 6.5×10^-9>4.75×10^-9.

Eureka Math Grade 8 Module 1 Lesson 13 Problem Set Answer Key

Question 1.
Write out a detailed proof of the fact that, given two numbers in scientific notation, a×10ⁿ and b×10ⁿ, a<b, if and only if a×10ⁿ<b×10ⁿ.
Answer:
Because 10ⁿ>0, we can use inequality (A) (i.e., (1) above) twice to draw the necessary conclusions. First, if a<b, then by inequality (A), a×10ⁿ<b×10ⁿ. Second, given a×10ⁿ<b×10ⁿ, we can use inequality (A) again to show an<b×10ⁿ by 10^-n.

a. Let A and B be two positive numbers, with no restrictions on their size. Is it true that A×10^-5<B×10⁵?
Answer:
No, it is not true that A×10^-5<B×10⁵. Using inequality (A), we can write
A×10^-5×10⁵<B×10⁵×10⁵, which is the same as A<B×10¹⁰. To disprove the statement, all we would need to do is find a value of A that exceeds B×10¹⁰.

b. Now, if A×10^-5 and B×10⁵ are written in scientific notation, is it true that A×10^-5<B×10⁵? Explain.
Answer:
Yes, since the numbers are written in scientific notation, we know that the restrictions for A and B are
1≤A<10 and 1≤B<10. The maximum value for A, when multiplied by 10^-5, will still be less than 1. The minimum value of B will produce a number at least 10⁵ in size.

Question 2.
The mass of a neutron is approximately 1.674927×10^-27 kg. Recall that the mass of a proton is
1.672622×10^-27 kg. Explain which is heavier.
Answer:
Since both numbers have a factor of 10^-27, we only need to look at 1.674927 and 1.672622. When we multiply each number by 10⁶, we get
1.674927×10⁶ and 1.672622×10⁶,
which is the same as
1,674,927 and 1,672,622.
Now that we are looking at whole numbers, we can see that 1,674,927>1,672,622 (by (2b) above), which means that 1.674927×10^-27>1.672622×10^-27. Therefore, the mass of a neutron is heavier.

Question 3.
The average lifetime of the Z boson is approximately 3×10^-25 seconds, and the average lifetime of a neutral rho meson is approximately 4.5×10^-24 seconds.
a. Without using the theorem from today’s lesson, explain why the neutral rho meson has a longer average lifetime.
Answer:
Since 3×10^-25=3×10^-1×10^-24, we can compare 3×10^-1×10^-24 and 4.5×10^-24. Based on Example 3 or by use of (1) above, we only need to compare 3×10^-1 and 4.5, which is the same as 0.3 and 4.5. If we multiply each number by 10, we get whole numbers 3 and 45. Since 3<45, then
3×10^-25<4.5×10^-24. Therefore, the neutral rho meson has a longer average lifetime.

b. Approximately how much longer is the lifetime of a neutral rho meson than a Z boson?
Answer:
45:3 or 15 times longer

Rapid White Board Exchange: Operations with Numbers Expressed in Scientific Notation

Question 1.
(5×10⁴)²=
Answer:
2.5×10⁹

Question 2.
(2×10⁹)⁴=
Answer:
1.6×10³7

Question 3.
\(\frac{\left(1.2 \times 10^{4}\right)+\left(2 \times 10^{4}\right)+\left(2.8 \times 10^{4}\right)}{3}\)=
Answer:
2×10⁴

Question 4.
\(\frac{7 \times 10^{15}}{14 \times 10^{9}}\) =
Answer:
5×10⁵

Question 5.
\(\frac{4 \times 10^{2}}{2 \times 10^{8}}\)=
Answer:
2×10^-6

Question 6.
\(\frac{\left(7 \times 10^{9}\right)+\left(6 \times 10^{9}\right)}{2}\)=
Answer:
6.5×10⁹

Question 7.
(9×10^-4)²=
Answer:
8.1×10^-7

Question 8.
(9.3×10¹⁰)-(9×10¹⁰)=
Answer:
3×10⁹

Engage NY Eureka Math 8th Grade Module 1 Lesson 12 Answer Key

Eureka Math Grade 8 Module 1 Lesson 12 Exercise Answer Key

Exercise 1.
A certain brand of MP3 player will display how long it will take to play through its entire music library. If the maximum number of songs the MP3 player can hold is 1,000 (and the average song length is 4 minutes), would you want the time displayed in terms of seconds-, days-, or years-worth of music? Explain.
Answer:
It makes the most sense to have the time displayed in days because numbers such as 240,000 seconds-worth of music and \(\frac{5}{657}\) of a year are more difficult to understand than about 2.8 days.

Exercise 2.
You have been asked to make frosted cupcakes to sell at a school fundraiser. Each frosted cupcake contains about 20 grams of sugar. Bake sale coordinators expect 500 people will attend the event. Assume everyone who attends will buy a cupcake; does it make sense to buy sugar in grams, pounds, or tons? Explain.
Answer:
Because each cupcake contains about 20 grams of sugar, we will need 500×20 grams of sugar. Therefore, grams are too small of a measurement, while tons are too large. Therefore, the sugar should be purchased in pounds.

Exercise 3.
The seafloor spreads at a rate of approximately 10 cm per year. If you were to collect data on the spread of the seafloor each week, which unit should you use to record your data? Explain.
Answer:
The seafloor spreads 10 cm per year, which is less than 1 cm per month. Data will be collected each week, so it makes the most sense to measure the spread with a unit like millimeters.

Exercise 4.
Show that the mass of a proton is 0.938 \(\frac{\mathrm{GeV}}{c^{2}}\).
Answer:
Let x represent the number of gigaelectronvolts equal to the mass of a proton.
x(\(\frac{\mathrm{GeV}}{c^{2}}\))=mass of proton
x(1.783×10^-27)=1.672622×10^-27
x=(1.672622×10^-27)/(1.783×10^-27)
=\(\frac{1.672622}{1.783}\)
≈0.938

Exercise 5.
The distance of the nearest star (Proxima Centauri) to the sun is approximately 4.013 336 473×10¹³ km. Show that Proxima Centauri is 4.2421 light-years from the sun.
Answer:
Let x represent the number of light-years Proxima Centauri is from the sun.
x(9.46073×10¹²)=4.013336473×10¹³
x=\(\frac{4.013336473 \times 10^{13}}{9.46073 \times 10^{12}}\)
=\(\frac{4.013336473}{9.46073}\)×10
=0.424210021×10
≈4.2421

Exercise 6.
What new unit might you introduce in order to discuss the differences in diameter measurements?
Answer:
There are several answers that students could give for their choice of unit. Accept any reasonable answer, provided the explanation is clear and correct. Some students may choose 10^-12 as their unit because all measurements could then be expressed without exponential notation. Other students may decide that 10^-8 should be the unit because two measurements are already of that order of magnitude. Still, other students may choose 10^-10 because that is the average of the powers.

Exercise 7.
Name your unit, and explain why you chose it.
Answer:
Students can name their unit anything reasonable, as long as they clearly state what their unit is and how it will be written. For example, if a student chooses a unit of 10^-10, then he or she should state that the unit will be represented with a letter. For example, Y, then Y=10^-10.

Exercise 8.
Using the unit you have defined, rewrite the five diameter measurements.
Answer:
Using the unit Y=10^-10, then:

Eureka Math Grade 8 Module 1 Lesson 12 Exploratory Challenge Answer Key

Exploratory Challenge 1.
Finally, let us look at an example involving the masses of the planets in our solar system. They range from Mercury’s 3.3022×10²³ kg to Jupiter’s 1.8986×10²⁷ kg. However, Earth’s mass is the fourth heaviest among the eight planets³, and it seems reasonable to use it as the point of reference for discussions among planets. Therefore, a new unit is M_E, the mass of the Earth, or 5.97219×10²⁴ kg.
Suggested white-board activity: Show students the table below, leaving the masses for Mercury and Jupiter blank. Demonstrate how to rewrite the mass of Mercury in terms of the new unit, M_E. Then, have students rewrite the mass of Jupiter using the new unit. Finally, complete the chart with the rewritten masses.

Mercury: Let x represent the mass of Mercury in the unit M_E. We want to determine what number times the new unit is equal to the mass of Mercury in kilograms. Since M_E=5.97219×10²⁴, then:
(5.97219×10²⁴ )x=3.3022×10²³
x=\(\frac{3.3022 \times 10^{23}}{5.97219 \times 10^{24}}\)
= \(\frac{3.3022}{5.97219}\) × \(\frac{10^{23}}{10^{24}}\)
≈0.553×10^1-
=0.0553.
Mercury’s mass is 0.0553 M_E.

Jupiter: Let x represent the mass of Jupiter in the unit M_E. We want to determine what number times the new unit is equal to the mass of Jupiter in kilograms. Since M_E=5.97219×10²⁴, then:
(5.97219×10²⁴ )x=1.8986×10²⁷
x=\(\frac{1.8986 \times 10^{27}}{5.97219 \times 10^{24}}\)
=\(\frac{1.8986}{5.97219}\)×\(\frac{10^{27}}{10^{24}}\)
≈0.318×10³
=318.
Jupiter’s mass is 318 M_E.

Exploratory Challenge 2.
Suppose you are researching atomic diameters and find that credible sources provided the diameters of five different atoms as shown in the table below. All measurements are in centimeters.

Eureka Math Grade 8 Module 1 Lesson 12 Exit Ticket Answer Key

Question 1.
The table below shows an approximation of the national debt at the beginning of each decade over the last century. Choose a unit that would make a discussion about the growth of the national debt easier. Name your unit, and explain your choice.

Answer:
Students will likely choose 10¹¹ as their unit because the majority of the data is of that magnitude. Accept any reasonable answer that students provide. Verify that they have named their unit.

Question 2.
Using the new unit you have defined, rewrite the debt for years 1900, 1930, 1960, and 2000.
Answer:
Let D represent the unit 10¹¹. Then, the debt in 1900 is 0.021D, in 1930 it is 0.16D, in 1960 it is 2.9D, and 57D in 2000.

Eureka Math Grade 8 Module 1 Lesson 12 Problem Set Answer Key

Question 1.
Verify the claim that, in terms of gigaelectronvolts, the mass of an electron is 0.000511.
Answer:
Let x represent the number of gigaelectronvolts equal to the mass of an electron.

Question 2.
The maximum distance between Earth and the sun is 1.52098232×10⁸ km, and the minimum distance is 1.47098290×10⁸ km. What is the average distance between Earth and the sun in scientific notation?
Answer:

Question 3.
Suppose you measure the following masses in terms of kilograms:

What new unit might you introduce in order to aid discussion of the masses in this problem? Name your unit, and express it using some power of 10. Rewrite each number using your newly defined unit.
Answer:
A very motivated student may search the Internet and find that units exist that convert large masses to reasonable numbers, such as petagrams (10¹² kg), exagrams (10¹⁵ kg), or zetagrams (10¹⁸ kg). More likely, students will decide that something near 10²⁰ should be used as a unit because many of the numbers are near that magnitude. There is one value, 1.8×10¹², that serves as an outlier and should be ignored because it is much smaller than the majority of the data. Students can name their unit anything reasonable. The answers provided are suggestions, but any reasonable answers should be accepted.
Let U be defined as the unit 10²⁰.

Engage NY Eureka Math 8th Grade Module 1 Lesson 11 Answer Key

Eureka Math Grade 8 Module 1 Lesson 11 Example Answer Key

Example 1.
The advantage of the scientific notation becomes even more pronounced when we have to compute how many times heavier a proton is than an electron. Instead of writing the value of the ratio, r, as

we express it as
r=\(\frac{1.672622 \times 10^{-27}}{9.10938291 \times 10^{-31}}\)

→ Should we eliminate the power of 10 in the numerator or denominator? Explain.
→ Using the theorem on generalized equivalent fractions, we can eliminate the negative power of 10 in the numerator and denominator to see what we are doing more clearly. Anticipating that 10^-31×10³¹=1, we can multiply the numerator and denominator of the (complex) fraction by 10³¹

Using the first law of exponents (10) presented in Lesson 5, we get

Note that since we are using scientific notation, we can interpret an approximate value of r right away. For example, we see

so that r is approximately \(\frac{1}{5}\) ×10,000, which is 2,000. Thus, we expect a proton to be about two thousand times heavier than an electron.

Example 2.
The U.S. national debt as of March 23, 2013, rounded to the nearest dollar, is $16,755,133,009,522. According to the 2012 U.S. census, there are about 313,914,040 U.S. citizens. What is each citizen’s approximate share of the debt?
\(\frac{1.6755 \times 10^{13}}{3.14 \times 10^{8}}\) = \(\frac{1.6755}{3.14}\) × \(\frac{10^{13}}{10^{8}}\)
=\(\frac{1.6755}{3.14}\)×10⁵
=0.533598…×10⁵
≈0.5336×10⁵
=53360
Each U.S. citizen’s share of the national debt is about $53,360.

Eureka Math Grade 8 Module 1 Lesson 11 Exercise Answer Key

Exercise 1.
The mass of a proton is
0.000 000 000 000 000 000 000 000 001 672 622 kg.
In scientific notation it is
Answer:
1.672622×10^-27 kg.

Exercise 2.
The mass of an electron is
0.000 000 000 000 000 000 000 000 000 000 910 938 291 kg.
In scientific notation it is
Answer:
9.10938291×10^-31 kg.

Exercise 3.
Write the ratio that compares the mass of a proton to the mass of an electron.
Answer:
Ratio: (1.672622×10^-27 ) : (9.10938291×10^-31)

Exercise 4.
Compute how many times heavier a proton is than an electron (i.e., find the value of the ratio). Round your final answer to the nearest one.
Answer:
Let r= the value of the ratio, then:

Exercise 5.
The geographic area of California is 163,696 sq.mi., and the geographic area of the U.S. is 3,794,101 sq.mi. Let’s round off these figures to 1.637×10⁵ and 3.794×10⁶. In terms of area, roughly estimate how many Californias would make up one U.S. Then compute the answer to the nearest ones.
Answer:
\(\frac{3.794 \times 10^{6}}{1.637 \times 10^{5}}\) = \(\frac{3.794}{1.637}\) × \(\frac{10^{6}}{10^{5}}\)
= \(\frac{3.794}{1.637}\)×10
=2.3176…×10
≈2.318×10
=23.18
It would take about 23 Californias to make up one U.S.

Exercise 6.
The average distance from Earth to the moon is about 3.84×10⁵ km, and the distance from Earth to Mars is approximately 9.24×10⁷ km in year 2014. On this simplistic level, how much farther is traveling from Earth to Mars than from Earth to the moon?
Answer:
9.24×10⁷-3.84×10⁵=924×10⁵-3.84×10⁵
=(924-3.84)×10⁵
=920.16×10⁵
=92 016 000
It is 92,016,000 km further to travel from Earth to Mars than from Earth to the moon.

Eureka Math Grade 8 Module 1 Lesson 11 Problem Set Answer Key

Question 1.
There are approximately 7.5×10¹⁸ grains of sand on Earth. There are approximately 7×10²⁷ atoms in an average human body. Are there more grains of sand on Earth or atoms in an average human body? How do you know?
Answer:
There are more atoms in the average human body. When comparing the order of magnitude of each number,
27>18; therefore, 7×10²⁷>7.5×10¹⁸.

Question 2.
About how many times more atoms are in a human body compared to grains of sand on Earth?
Answer:
\(\frac{7 \times 10^{27}}{7.5 \times 10^{18}}\) = \(\frac{7}{7.5}\)×\(\frac{10^{27}}{10^{18}}\)
≈1×10^27-18
≈1×10⁹
≈10⁹
There are about 1,000,000,000 times more atoms in the human body compared to grains of sand on Earth.

Question 3.
Suppose the geographic areas of California and the U.S. are 1.637×10⁵ and 3.794×10⁶ sq.mi., respectively. California’s population (as of 2012) is approximately 3.804×10⁷ people. If population were proportional to area, what would be the U.S. population?
Answer:
We already know from Exercise 5 that it would take about 23 Californias to make up one U.S. Then the population of the U.S. would be 23 times the population of California, which is
23×3.804×10⁷=87.492×10⁷
=8.7492×10⁸
=874,920,000.

Question 4.
The actual population of the U.S. (as of 2012) is approximately 3.14×10⁸. How does the population density of California (i.e., the number of people per square mile) compare with the population density of the U.S.?
Answer:
Population density of California per square mile:
\(\frac{3.804 \times 10^{7}}{1.637 \times 10^{5}}\) = \(\frac{3.804}{1.637}\) ×\(\frac{10^{7}}{10^{5}}\)
=2.32376…×10²
≈2.32×10²
=232
Population density of the U.S. per square mile:
\(\frac{3.14 \times 10^{8}}{3.794 \times 10^{6}}\) = \(\frac{3.14}{3.794}\)×\(\frac{10^{8}}{10^{6}}\)
≈0.83×10²
=83
Population density of California compared to the population density of the U.S.:
\(\frac{232}{83}\)=2.7951…
≈2.8
California is about 3 times as dense as the U.S. in terms of population.

Eureka Math Grade 8 Module 1 Lesson 11 Exit Ticket Answer Key

Question 1.
Two of the largest mammals on earth are the blue whale and the elephant. An adult male blue whale weighs about 170 tonnes or long tons. (1 tonne = 1000 kg)
Show that the weight of an adult blue whale is 1.7×10⁵ kg.
Answer:
Let x (or any other symbol) represent the number of kilograms an adult blue whale weighs.
170×1000=x
1.7×10⁵=x

Question 2.
An adult male elephant weighs about 9.07×10³ kg.
Compute how many times heavier an adult male blue whale is than an adult male elephant (i.e., find the value of the ratio). Round your final answer to the nearest one.
Answer:
Let r be the value of the ratio.
r=\(\frac{1.7 \times 10^{5}}{9.07 \times 10^{3}}\)
=\(\frac{1.7}{9.07}\)×10²
=0.18743×10²
=18.743
≈19
The blue whale is 19 times heavier than the elephant.

Engage NY Eureka Math 8th Grade Module 1 Lesson 10 Answer Key

Eureka Math Grade 8 Module 1 Lesson 10 Exercise Answer Key

Exercise 1.
The speed of light is 300,000,000 meters per second. The sun is approximately 1.5×10¹¹ meters from Earth. How many seconds does it take for sunlight to reach Earth?
Answer:
300 000 000=3×10⁸
\(\frac{1.5 \times 10^{11}}{3 \times 10^{8}}\) = \(\frac{1.5}{3}\)×\(\frac{10^{11}}{10^{8}}\)
=0.5×10³
=0.5×10×10²
=5×10²
It takes 500 seconds for sunlight to reach Earth.

Exercise 2.
The mass of the moon is about 7.3×10²² kg. It would take approximately 26,000,000 moons to equal the mass of the sun. Determine the mass of the sun.
Answer:
26 000 000=2.6×10⁷
(2.6× 10⁷)(7.3×10²²)=(2.6×7.3)(10⁷×10²²)
=18.98×10²⁹
=1.898×10×10²⁹
=1.898×10³⁰
The mass of the sun is 1.898×10³⁰ kg.

Exercise 3.
The mass of Earth is 5.9×10²⁴ kg. The mass of Pluto is 13,000,000,000,000,000,000,000 kg. Compared to Pluto, how much greater is Earth’s mass than Pluto’s mass?
Answer:
13 000 000 000 000 000 000 000=1.3×10²²
5.9×10²⁴-1.3×10²²=(5.9×10²)×10²²-1.3×10²²
=(590-1.3)×10²²
=588.7×10²²
=5.887×10²×10²²
=5.887×10²⁴
The mass of Earth is 5.887×10²⁴ kg greater than the mass of Pluto.

Exercise 4.
Using the information in Exercises 2 and 3, find the combined mass of the moon, Earth, and Pluto.
Answer:
7.3×10²²+1.3×10²²+5.9×10²⁴=(7.3×10²²+1.3×10²²)+5.9×10²⁴
=8.6×10²²+5.9×10²⁴
=(8.6+590)×10²²
=598.6×10²²
=5.986×10²×10²²
=5.986×10²⁴
The combined mass of the moon, Earth, and Pluto is 5.986×10²⁴ kg.

Exercise 5.
How many combined moon, Earth, and Pluto masses (i.e., the answer to Exercise 4) are needed to equal the mass of the sun (i.e., the answer to Exercise 2)?
Answer:
\(\frac{1.898 \times 10^{30}}{5.986 \times 10^{24}}\) = \(\frac{1.898}{5.986}\) × \(\frac{10^{30}}{10^{24}}\)
=0.3170…×10⁶
≈0.32×10⁶
=0.32×10×10⁵
=3.2×10⁵
It would take 3.2×10⁵ combined masses of the moon, Earth, and Pluto to equal the mass of the sun.

Eureka Math Grade 8 Module 1 Lesson 10 Problem Set Answer Key

Have students practice operations with numbers written in scientific notation and standard notation.

Question 1.
The sun produces 3.8×10²7 joules of energy per second. How much energy is produced in a year? (Note: a year is approximately 31,000,000 seconds).
Answer:
31 000 000=3.1×10⁷
(3.8×10²⁷)(3.1×10⁷)=(3.8×3.1)(10²⁷×10⁷)
=11.78×10³⁴
=1.178×10×10³⁴
=1.178×10³⁵
The sun produces 1.178×10³⁵ joules of energy in a year.

Question 2.
On average, Mercury is about 57,000,000 km from the sun, whereas Neptune is about 4.5×10⁹ km from the sun. What is the difference between Mercury’s and Neptune’s distances from the sun?
Answer:
57 000 000=5.7×10⁷
4.5×10⁹-5.7×10⁷=(4.5×10²)×10⁷-5.7×10⁷
=450×10⁷-5.7×10⁷
=(450-5.7)×10⁷
=444.3×10⁷
=4.443×10²×10⁷
=4.443×10⁹
The difference in the distance of Mercury and Neptune from the sun is 4.443×10⁹ km.

Question 3.
The mass of Earth is approximately 5.9×10²⁴ kg, and the mass of Venus is approximately 4.9×10²⁴ kg.
a. Find their combined mass.
Answer:
5.9×10²⁴+4.9×10²⁴=(5.9+4.9)×10²⁴
=10.8×10²⁴
=1.08×10×10²⁴
=1.08×10²⁵
The combined mass of Earth and Venus is 1.08×10²⁵ kg.

b. Given that the mass of the sun is approximately 1.9×10³⁰ kg, how many Venuses and Earths would it take to equal the mass of the sun?
Answer:
\(\frac{1.9 \times 10^{30}}{1.08 \times 10^{25}}\)=\(\frac{1.9}{1.08} \times \frac{10^{30}}{10^{25}}\)
=1.75925…×10⁵
≈1.8×10⁵
It would take approximately 1.8×10⁵ Venuses and Earths to equal the mass of the sun.

Eureka Math Grade 8 Module 1 Lesson 10 Exit Ticket Answer Key

Question 1.
The speed of light is 3×10⁸ meters per second. The sun is approximately 230,000,000,000 meters from Mars. How many seconds does it take for sunlight to reach Mars?
Answer:
230 000 000 000=2.3×10¹¹
\(\frac{2.3 \times 10^{11}}{3 \times 10^{8}}\)=\(\frac{2.3}{3}\)×\(\frac{10^{11}}{10^{8}}\)
=0.7666…×10³
≈0.77×10×10²
≈7.7×10²
It takes approximately 770 seconds for sunlight to reach Mars.

Question 2.
If the sun is approximately 1.5×10¹¹ meters from Earth, what is the approximate distance from Earth to Mars?
Answer:
(2.3×10¹¹)-(1.5×10¹¹)=(2.3-1.5)×10¹¹
=0.8×10¹¹
=0.8×10×10¹⁰
=8×10¹⁰
The distance from Earth to Mars is 8×10¹⁰ meters.

Engage NY Eureka Math 8th Grade Module 6 End of Module Assessment Answer Key

Eureka Math Grade 8 Module 6 End of Module Assessment Task Answer Key

Question 1.
The Kentucky Derby is a horse race held each year. The following scatter plot shows the speed of the winning horse at the Kentucky Derby each year between 1875 and 2012.

Data Source: http://www.kentuckyderby.com/
(Note: Speeds were calculated based on times given on website.)
a. Is the association between speed and year positive or negative? Give a possible explanation in the context of this problem for why the association behaves this way considering the variables involved.
Answer:
The association is positive overall, as horses have been getting faster over time. This is perhaps due to improved training methods.

b. Comment on whether the association between speed and year is approximately linear, and then explain in the context of this problem why the form of the association (linear or not) makes sense considering the variables involved.
Answer:
The association is not linear. There is probably a physical limit to how fast horses can go that we are approaching.

c. Circle an outlier in this scatter plot, and explain, in context, how and why the observation is unusual.
Answer:
The winner that year was much slower than we could have predicted.

Question 2.
Students were asked to report their gender and how many times a day they typically wash their hands. Of the 738 males, 66 said they wash their hands at most once a day, 583 said two to seven times per day, and 89 said eight or more times per day. Of the 204 females, 2 said they wash their hands at most once a day, 160 said two to seven times per day, and 42 said eight or more times per day.
a. Summarize these data in a two-way table with rows corresponding to the three different frequency-of-hand-washing categories and columns corresponding to gender.
Answer:

b. Do these data suggest an association between gender and frequency of hand washing? Support your answer with appropriate calculations.
Answer:

Males are more likely than females to wash hands at most once per day. Females are more likely to wash 8 or more times per day.

Question 3.
Basketball players who score a lot of points also tend to be strong in other areas of the game such as number of rebounds, number of blocks, number of steals, and number of assists. Below are scatter plots and linear models for professional NBA (National Basketball Association) players last season.

a. The line that models the association between points scored and number of rebounds is y = 21.54 + 3.833x, where y represents the number of points scored and x represents the number of rebounds. Give an interpretation, in context, of the slope of this line.
Answer:
If the number of rebounds increases by one, we predict the number of points increases by 3.833.

b. The equations on the previous page all show the number of points scored (y) as a function of the other variables. An increase in which of the variables (rebounds, blocks, steals, and assists) tends to have the largest impact on the predicted points scored by an NBA player?
Answer:
Each additional block corresponds to 22.45 more points, the largest slope or rate of increase.

c. Which of the four linear models shown in the scatter plots on the previous page has the worst fit to the data? Explain how you know using the data.
Answer:
Probably number of blocks because the association is weaker. There is more scatter of the points away from the line.

Engage NY Eureka Math 8th Grade Module 6 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 6 Mid Module Assessment Task Answer Key

Question 1.
Many computers come with a Solitaire card game. The player moves cards in certain ways to complete specific patterns. The goal is to finish the game in the shortest number of moves possible, and a player’s score is determined by the number of moves. A statistics teacher played the game 16 times and recorded the number of moves and the final score after each game. The line represents the linear function that is used to determine the score from the number of moves.

a. Was this person’s average score closer to 1130 or 1110? Explain how you decided.
Answer:
Most of the games had scores between 1125 and 1175. The mean score will be closer to 1130.

b. The first two games she played took 169 moves (1131 points) and 153 moves (1147 points). Based on this information, determine the equation of the linear function used by the computer to calculate the score from the number of moves. Explain your work.
Answer:
The difference in the scores is 1131 – 1147 or – 16.
The difference in the number of moves is 169 – 153 = 16
The slope is –\(\frac{16}{16}\) or – 1. This means that 1131 = intercept – 169, So intercept equals 1300
Score = 1300 – moves

c. Based on the linear function, each time the player makes a move, how many points does she lose?
Answer:
One point last per move.

d. Based on the linear function, how many points does the player start with in this game? Explain your reasoning.
Answer:
1300, or the score when the number of moves equals 0.

Question 2.
To save money, drivers often try to increase their mileage, which is measured in miles per gallon (mpg). One theory is that speed traveled impacts mileage. Suppose the following data are recorded for five different 300-mile tests, with the car traveling at different speeds in miles per hour (mph) for each test.

a. For the data in this table, is the association positive or negative? Explain how you decided.
Answer:
As the speed increases in miles per hour, the miles per gallon decrease. This describes a negative association.

b. Construct a scatter plot of these data using the following coordinate grid. The vertical axis represents the mileage, and the horizontal axis represents the speed in miles per hour (mph).

Answer:

c. Draw a line on your scatter plot that you think is a reasonable model for predicting the mileage from the car speed.
Answer:

d. Estimate and interpret the slope of the line you found in part (c).
Answer:
Two points are approximately (80, 20) and (50, 33).
So, slope ≈ \(\frac{20-33}{80-50}\) ≈ – \(0.43 \overline{3}\)
Each increase of 1 mph in speed predicts a decrease of \(0.43 \overline{3}\) mpg.

Suppose additional data were measured for three more tests. These results have been added to the previous tests, and the combined data are shown in the table below.

e. Does the association for these data appear to be linear? Why or why not?
Answer:
No, While the speeds increase, the milage value increase and then mostly decrease. There is no fixed rate of increase or decrease for milage based on increased speed.

f. If your only concern was mileage and you had no traffic constraints, what speed would you recommend traveling based on these data? Explain your choice.
Answer:
About 50 mph. It is around 50 mph that the mpg stops increasing and starts to decrease.

Engage NY Eureka Math 8th Grade Module 6 Lesson 14 Answer Key

Eureka Math Grade 8 Module 6 Lesson 14 Exercise Answer Key

Example 1.
Suppose a random group of people are surveyed about their use of smartphones. The results of the survey are summarized in the tables below.

Example 2.
Suppose a sample of 400 participants (teachers and students) was randomly selected from the middle schools and high schools in a large city. These participants responded to the following question:

Which type of movie do you prefer to watch?
1. Action (The Avengers, Man of Steel, etc.)
2. Drama (42 (The Jackie Robinson Story), The Great Gatsby, etc.)
3. Science Fiction (Star Trek into Darkness, World War Z, etc.)
4. Comedy (Monsters University, Despicable Me 2, etc.)

Movie preference and status (teacher or student) were recorded for each participant.

Exercises 1–7

Exercise 1.
Two variables were recorded. Are these variables categorical or numerical?
Answer:
Both variables are categorical.

Exercise 2.
The results of the survey are summarized in the table below.

a. What proportion of participants who are teachers prefer action movies?
Answer:
\(\frac{40}{100}\) = 0.40

b. What proportion of participants who are teachers prefer drama movies?
Answer:
\(\frac{20}{100}\) = 0.20

c. What proportion of participants who are teachers prefer science fiction movies?
Answer:
\(\frac{10}{100}\) = 0.10

d. What proportion of participants who are teachers prefer comedy movies?
Answer:
\(\frac{30}{100}\) = 0.30

The answers to Exercise 2 are called row relative frequencies. Notice that you divided each cell frequency in the Teacher row by the total for that row. Below is a blank relative frequency table.
Table of Row Relative Frequencies

Answer:

Write your answers from Exercise 2 in the indicated cells in the table above.

Exercise 3.
Find the row relative frequencies for the Student row. Write your answers in the table above.
a. What proportion of participants who are students prefer action movies?
b. What proportion of participants who are students prefer drama movies?
c. What proportion of participants who are students prefer science fiction movies?
d. What proportion of participants who are students prefer comedy movies?
Answer:
See the table above.

Exercise 4.
Is a participant’s status (i.e., teacher or student) related to what type of movie he would prefer to watch? Why or why not? Discuss this with your group.
Answer:
No. Teachers are just as likely to prefer each movie type as students are, according to the row relative frequencies.

Exercise 5.
What does it mean when we say that there is no association between two variables? Discuss this with your group.
Answer:
Answers will vary. No association means that knowing the value of one variable does not tell anything about the value of the other variable.

Exercise 6.
Notice that the row relative frequencies for each movie type are the same for both the Teacher and Student rows. When this happens, we say that the two variables, movie preference and status (student or teacher), are not associated. Another way of thinking about this is to say that knowing if a participant is a teacher (or a student) provides no information about his movie preference.
What does it mean if row relative frequencies are not the same for all rows of a two-way table?
Answer:
It means that there is an association or a tendency between the two variables.

Exercise 7.
You can also evaluate whether two variables are associated by looking at column relative frequencies instead of row relative frequencies. A column relative frequency is a cell frequency divided by the corresponding column total.
For example, the column relative frequency for the Student/Action cell is \(\frac{120}{160}\) = 0.75.
a. Calculate the other column relative frequencies, and write them in the table below.
Table of Column Relative Frequencies

Answer:
Table of Column Relative Frequencies

b. What do you notice about the column relative frequencies for the four columns?
Answer:
The column relative frequencies are equal for all four columns.

c. What would you conclude about association based on the column relative frequencies?
Answer:
Because the column relative frequencies are the same for all four columns, we would conclude that there is no association between movie preference and status.

Example 3
In the survey described in Example 2, gender for each of the 400 participants was also recorded. Some results of the survey are given below:

• 160 participants preferred action movies.
• 80 participants preferred drama movies.
• 40 participants preferred science fiction movies.
• 240 participants were females.
• 78 female participants preferred drama movies.
• 32 male participants preferred science fiction movies.
• 60 female participants preferred action movies.

Exercises 8–15
Use the results from Example 3 to answer the following questions. Be sure to discuss these questions with your group members.

Exercise 8.
Complete the two-way frequency table that summarizes the data on movie preference and gender.

Answer:

Exercise 9.
What proportion of the participants are female?
Answer:
\(\frac{240}{400}\) = 0.60

Exercise 10.
If there was no association between gender and movie preference, should you expect more females than males or fewer females than males to prefer action movies? Explain.
Answer:
If there was no association between gender and movie preference, then I would expect more females than males to prefer action movies just because there are more females in the sample. However, if there was an association between gender and movie preference, then I would expect either fewer females than males who prefer action movies or considerably more females than males who prefer action movies.

Exercise 11.
Make a table of row relative frequencies of each movie type for the Male row and the Female row. Refer to Exercises 2–4 to review how to complete the table below.

Answer:

Suppose that you randomly pick 1 of the 400 participants. Use the table of row relative frequencies on the previous page to answer the following questions.
Exercise 12.
If you had to predict what type of movie this person chose, what would you predict? Explain why you made this choice.
Answer:
The participant likely prefers action movies because the largest proportion of participants preferred action movies.

Exercise 13.
If you know that the randomly selected participant is female, would you predict that her favorite type of movie is action? If not, what would you predict, and why?
Answer:
No. A female participant is more likely to prefer comedy since it has the greatest row relative frequency in the Female row.

Exercise 14.
If knowing the value of one of the variables provides information about the value of the other variable, then there is an association between the two variables.
Is there an association between the variables gender and movie preference? Explain.
Answer:
Yes. The row relative frequencies are not the same (not even close) in each row in the table.

Exercise 15.
What can be said when two variables are associated? Read the following sentences. Decide if each sentence is a correct statement based upon the survey data. If it is not correct, explain why not.
a. More females than males participated in the survey.
Answer:
Correct

b. Males tend to prefer action and science fiction movies.
Answer:
Correct

c. Being female causes one to prefer drama movies.
Answer:
Incorrect Association does not imply a cause-and-effect relationship.

Eureka Math Grade 8 Module 6 Lesson 14 Problem Set Answer Key

A sample of 200 middle school students was randomly selected from the middle schools in a large city. Answers to several survey questions were recorded for each student. The tables below summarize the results of the survey.
For each table, calculate the row relative frequencies for the Female row and for the Male row. Write the row relative frequencies beside the corresponding frequencies in each table below.

Question 1.
This table summarizes the results of the survey data for the two variables, gender and which sport the students prefer to play. Is there an association between gender and which sport the students prefer to play? Explain.

Answer:

Yes, there appears to be an association between gender and sports preference. The row relative frequencies are not the same for the Male and the Female rows, as shown in the table above.

Question 2.
This table summarizes the results of the survey data for the two variables, gender and the students’ T-shirt sizes. Is there an association between gender and T-shirt size? Explain.

Answer:

Yes, there appears to be an association between gender and T-shirt size. The row relative frequencies are not the same for the Male and the Female rows, as shown in the table above.

Question 3.
This table summarizes the results of the survey data for the two variables, gender and favorite type of music. Is there an association between gender and favorite type of music? Explain.

Answer:

No, there does not appear to be an association between gender and favorite type of music. The row relative frequencies are about the same for the Male and Female rows, as shown in the table above.

Eureka Math Grade 8 Module 6 Lesson 14 Exit Ticket Answer Key

A random sample of 100 eighth-grade students are asked to record two variables: whether they have a television in their bedrooms and if they passed or failed their last math test. The results of the survey are summarized below.

• 55 students have a television in their bedrooms.
• 35 students do not have a television in their bedrooms and passed their last math test.
• 25 students have a television and failed their last math test.
• 35 students failed their last math test.

Question 1.
Complete the two-way table.

Answer:

Question 2.
Calculate the row relative frequencies, and enter the values in the table above. Round to the nearest thousandth.
Answer:
The row relative frequencies are displayed in the table above.

Question 3.
Is there evidence of association between the variables? If so, does this imply there is a cause-and-effect relationship? Explain.
Answer:
Yes, there is evidence of association between the variables because the relative frequencies are different among the rows. However, this does not necessarily imply a cause-and-effect relationship. The fact that a student has a television in his room does not cause the student to fail a test. Rather, it may be that the student is spending more time watching television or playing video games instead of studying.

Engage NY Eureka Math 8th Grade Module 6 Lesson 12 Answer Key

Eureka Math Grade 8 Module 6 Lesson 12 Exercise Answer Key

Example 1: Growing Dahlias
A group of students wanted to determine whether or not compost is beneficial in plant growth. The students used the dahlia flower to study the effect of composting. They planted eight dahlias in a bed with no compost and another eight plants in a bed with compost. They measured the height of each plant over a 9-week period. They found the median growth height for each group of eight plants. The table below shows the results of the experiment for the dahlias grown in non-compost beds.

Exercises 1 – 15

Exercise 1.
On the grid below, construct a scatter plot of non-compost height versus week.

Answer:

Exercise 2.
Draw a line that you think fits the data reasonably well.
Answer:

Exercise 3.
Find the rate of change of your line. Interpret the rate of change in terms of growth (in height) over time.
Answer:
Most students should have a rate of change of approximately 3.5 inches per week. A rate of change of 3.5 means that the median height of the eight dahlias increased by about 3.5 inches each week.

Exercise 4.
Describe the growth (change in height) from week to week by subtracting the previous week’s height from the current height. Record the weekly growth in the third column in the table below. The median growth for the dahlias from Week 1 to Week 2 was 3.75 inches (i.e., 12.75-9.00=3.75).

Answer:

Exercise 5.
As the number of weeks increases, describe how the weekly growth is changing.
Answer:
The growth each week remains about the same—approximately 3.5 inches.

Exercise 6.
How does the growth each week compare to the slope of the line that you drew?
Answer:
The amount of growth per week varies from 3.25 to 3.75 but centers around 3.5, which is the slope of the line.

Exercise 7.
Estimate the median height of the dahlias at 8 \(\frac{1}{2}\) weeks. Explain how you made your estimate.
Answer:
An estimate is 35.5 inches. Students can use the graph, the table, or the equation of their line.

The table below shows the results of the experiment for the dahlias grown in compost beds.

Exercise 8.
Construct a scatter plot of height versus week on the grid below.

Answer:

Exercise 9.
Do the data appear to form a linear pattern?
Answer:
No, the pattern in the scatter plot is curved.

Exercise 10.
Describe the growth from week to week by subtracting the height from the previous week from the current height. Record the weekly growth in the third column in the table below. The median weekly growth for the dahlias from Week 1 to Week 2 is 3.5 inches. (i.e., 13.5-10=3.5).

Answer:

Exercise 11.
As the number of weeks increases, describe how the growth changes.
Answer:
The amount of growth per week varies from week to week. In Weeks 1 through 4, the growth is around 4 inches each week. From Weeks 5 to 7, the amount of growth increases, and then the growth slows down for Weeks 8
and 9.

Exercise 12.
Sketch a curve through the data. When sketching a curve, do not connect the ordered pairs, but draw a smooth curve that you think reasonably describes the data.
Answer:

Exercise 13.
Use the curve to estimate the median height of the dahlias at 8 1\(\frac{1}{2}\) weeks. Explain how you made your estimate.
Answer:
Answers will vary. A reasonable estimate of the median height at 8 1\(\frac{1}{2}\) weeks is approximately 85 inches. Starting at 8 1\(\frac{1}{2}\) on the x-axis, move up to the curve and then over to the y-axis for the estimate of the height.

Exercise 14.
How does the weekly growth of the dahlias in the compost beds compare to the weekly growth of the dahlias in the non-compost beds?
Answer:
The growth in the non-compost is about the same each week. The growth in the compost starts the same as the non-compost, but after four weeks, the dahlias begin to grow at a faster rate.

Exercise 15.
When there is a car accident, how do the investigators determine the speed of the cars involved? One way is to measure the skid marks left by the cars and use these lengths to estimate the speed.
The table below shows data collected from an experiment with a test car. The first column is the length of the skid mark (in feet), and the second column is the speed of the car (in miles per hour).

Data Source: http://forensicdynamics.com/stopping-braking-distance-calculator
(Note: Data has been rounded.)
a. Construct a scatter plot of speed versus skid-mark length on the grid below.

Answer:

b. The relationship between speed and skid-mark length can be described by a curve. Sketch a curve through the data that best represents the relationship between skid-mark length and the speed of the car. Remember to draw a smooth curve that does not just connect the ordered pairs.
Answer:
See the plot above.

c. If the car left a skid mark of 60 ft., what is an estimate for the speed of the car? Explain how you determined the estimate.
Answer:
The speed is approximately 38 mph. Using the graph, for a skid mark of 65 ft., the speed was 40 mph, so the estimate is slightly less than 40 mph.

d. A car left a skid mark of 150 ft. Use the curve you sketched to estimate the speed at which the car was traveling.
Answer:
62.5 mph

e. If a car leaves a skid mark that is twice as long as another skid mark, was the car going twice as fast? Explain.
Answer:
No. When the skid mark was 105 ft. long, the car was traveling 50 mph. When the skid mark was 205 ft. long (about twice the 105 ft.), the car was traveling 70 mph, which is not twice as fast.

Eureka Math Grade 8 Module 6 Lesson 12 Problem Set Answer Key

Question 1.
Once the brakes of the car have been applied, the car does not stop immediately. The distance that the car travels after the brakes have been applied is called the braking distance. The table below shows braking distance (how far the car travels once the brakes have been applied) and the speed of the car.

Data Source: http://forensicdynamics.com/stopping-braking-distance-calculator
(Note: Data has been rounded.)
a. Construct a scatter plot of braking distance versus speed on the grid below.

Answer:

b. Find the amount of additional distance a car would travel after braking for each speed increase of 10 mph. Record your answers in the table below.

Answer:

c. Based on the table, do you think the data follow a linear pattern? Explain your answer.
Answer:
No. If the relationship is linear, the values in the Amount of Distance Increase column would be approximately equal.

d. Describe how the distance it takes a car to stop changes as the speed of the car increases.
Answer:
As the speed of the car increases, the distance it takes the car to stop also increases.

e. Sketch a smooth curve that you think describes the relationship between braking distance and speed.
Answer:

f. Estimate braking distance for a car traveling at 52 mph. Estimate braking distance for a car traveling at 75 mph. Explain how you made your estimates.
Answer:
For 52 mph, the braking distance is about 115 ft.
For 75 mph, the braking distance is about 230 ft.
Both estimates can be made by starting on the x-axis, moving up to the curve, and then moving over to the y-axis.

Question 2.
The scatter plot below shows the relationship between cost (in dollars) and radius length (in meters) of fertilizing different-sized circular fields. The curve shown was drawn to describe the relationship between cost and radius.

a. Is the curve a good fit for the data? Explain.
Answer:
Yes, the curve fits the data very well. The data points lie close to the curve.

b. Use the curve to estimate the cost for fertilizing a circular field of radius 30 m. Explain how you made your estimate.
Answer:
Using the curve drawn on the graph, the cost is approximately $200 – $250.

c. Estimate the radius of the field if the fertilizing cost was $2,500. Explain how you made your estimate.
Answer:
Using the curve, an estimate for the radius is approximately 94 m. Locate the approximate cost of $2,500. The approximate radius for that point is 94 m.

Question 3.
Suppose a dolphin is fitted with a GPS that monitors its position in relationship to a research ship. The table below contains the time (in seconds) after the dolphin is released from the ship and the distance (in feet) the dolphin is from the research ship.

Answer:

a. Construct a scatter plot of distance versus time on the grid below.

Answer:

b. Find the additional distance the dolphin traveled for each increase of 50 seconds. Record your answers in the table above.
Answer:
See the table above.

c. Based on the table, do you think that the data follow a linear pattern? Explain your answer.
Answer:
No, the change in distance from the ship is not constant.

d. Describe how the distance that the dolphin is from the ship changes as the time increases.
Answer:
As the time away from the ship increases, the distance the dolphin is from the ship is also increasing. The farther the dolphin is from the ship, the faster it is swimming.

e. Sketch a smooth curve that you think fits the data reasonably well.
Answer:

f. Estimate how far the dolphin will be from the ship after 180 seconds. Explain how you made your estimate.
Answer:
About 500 ft. Starting on the x-axis at approximately 180 seconds, move up to the curve and then over to the y-axis to find an estimate of the distance.

Eureka Math Grade 8 Module 6 Lesson 12 Exit Ticket Answer Key

The table shows the population of New York City from 1850 to 2000 for every 50 years.

Data Source: www.census.gov
Answer:

Question 1.
Find the growth of the population from 1850 to 1900. Write your answer in the table in the row for the year 1900.
Answer:
2,921,655

Question 2.
Find the growth of the population from 1900 to 1950. Write your answer in the table in the row for the year 1950.
Answer:
4,454,755

Question 3.
Find the growth of the population from 1950 to 2000. Write your answer in the table in the row for the year 2000.
Answer:
116,321

Question 4.
Does it appear that a linear model is a good fit for the data? Why or why not?
Answer:
No, a linear model is not a good fit for the data. The rate of population growth is not constant; the values in the change in population column are all different.

Question 5.
Describe how the population changes as the years increase.
Answer:
As the years increase, the population increases.

Question 6.
Construct a scatter plot of time versus population on the grid below. Draw a line or curve that you feel reasonably describes the data.

Answer:
Students should sketch a curve. If students use a straight line, point out that the line does not reasonably describe the data, as some of the data points are far away from the line.

Question 7.
Estimate the population of New York City in 1975. Explain how you found your estimate.
Answer:
It is approximately 8,000,000. An estimate can be found by recognizing that the growth of the city did not change very much from 1950 to 2000. The mean of the 1950 population and the 2000 population could also be found.

Engage NY Eureka Math 8th Grade Module 6 Lesson 13 Answer Key

Eureka Math Grade 8 Module 6 Lesson 13 Exercise Answer Key

Exercises 1–8
On an upcoming field day at school, the principal wants to provide ice cream during lunch. She offers three flavors: chocolate, strawberry, and vanilla. She selected your class to complete a survey to help her determine how much of each flavor to buy.

Exercise 1.
Answer the following question. Wait for your teacher to count how many students selected each flavor.
Then, record the class totals for each flavor in the table below.
“Which of the following three ice cream flavors is your favorite: chocolate, strawberry, or vanilla?”

Answer:
Answers will vary. One possibility is shown below.

Exercise 2.
Which ice cream flavor do most students prefer?
Answer:
Students should respond with the most-selected flavor. For the data set shown here, that is chocolate.

Exercise 3.
Which ice cream flavor do the fewest students prefer?
Answer:
Students should respond with the least-selected flavor. For the data set shown here, that is strawberry.

Exercise 4.
What percentage of students preferred each flavor? Round to the nearest tenth of a percent.
Answer:
Answers will vary based on the data gathered in Exercise 1.
Chocolate: \(\frac{17}{28}\) ≈ 60.7%
Strawberry: \(\frac{4}{28}\) ≈ 14.3%
Vanilla: \(\frac{7}{28}\) = 25%

Exercise 5.
Do the numbers in the table in Exercise 1 summarize data on a categorical variable or a numerical variable?
Answer:
The numbers in this table summarize data on a categorical variable—the preferred flavor of ice cream.

Exercise 6.
Do the students in your class represent a random sample of all the students in your school? Why or why not? Discuss this with your neighbor.
Answer:
No, because there is no indication that the students were selected randomly.

Exercise 7.
Is your class representative of all the other classes at your school? Why or why not? Discuss this with your neighbor.
Answer:
This class might be representative of the other eighth-grade classes but might not be representative of sixth- and seventh-grade classes.

Exercise 8.
Do you think the principal will get an accurate estimate of the proportion of students who prefer each ice cream flavor for the whole school using only your class? Why or why not? Discuss this with your neighbor.
Answer:
It is unlikely to give a good estimate. It would depend on how representative the class is of all of the students at the school.

Example 1
Students in a different class were asked the same question about their favorite ice cream flavors. The table below shows the ice cream flavors and the number of students who chose each flavor for that particular class. This table is called a one-way frequency table because it shows the counts of a univariate categorical variable.

We compute the relative frequency for each ice cream flavor by dividing the count by the total number of observations.
relative frequency = \(\frac{\text { count for a category }}{\text { total number of observations }}\)

Since 11 out of 25 students answered chocolate, the relative frequency would be \(\frac{11}{25}\) = 0.44. This relative frequency shows that 44% of the class prefers chocolate ice cream. In other words, the relative frequency is the proportional value that each category is of the whole.

Exercises 9–10
Use the table for the preferred ice cream flavors from the class in Example 1 to answer the following questions.

Exercise 9.
What is the relative frequency for the category strawberry?
Answer:
Relative frequency = \(\frac{4}{25}\) = 0.16

Exercise 10.
Write a sentence interpreting the relative frequency value in the context of strawberry ice cream preference.
Answer:
16% of the students in this class prefer strawberry ice cream.

Example 2.
The principal also wondered if boys and girls have different favorite ice cream flavors. She decided to redo the survey by taking a random sample of students from the school and recording both their favorite ice cream flavors and their genders. She asked the following two questions:
“Which of the following ice cream flavors is your favorite: chocolate, strawberry, or vanilla?”
“What is your gender: male or female?”

The results of the survey are as follows:
Of the 30 students who prefer chocolate ice cream, 22 are males.
Of the 25 students who prefer strawberry ice cream, 15 are females.
Of the 27 students who prefer vanilla ice cream, 13 are males.

The values of two variables, which were ice cream flavor and gender, were recorded in this survey. Since both of the variables are categorical, the data are bivariate categorical data.

Exercises 11–17

Exercise 11.
Can we display these data in a one-way frequency table? Why or why not?
Answer:
No, a one-way frequency table is for univariate data. Here we have bivariate data, so we would need to use a two-way table.

Exercise 12.
Summarize the results of the second survey of favorite ice cream flavors in the following table:

Answer:

Exercise 13.
Calculate the relative frequencies of the data in the table in Exercise 12, and write them in the following table.

Answer:

Use the relative frequency values in the table to answer the following questions:
Exercise 14.
What is the proportion of the students who prefer chocolate ice cream?
Answer:
0.37

Exercise 15.
What is the proportion of students who are female and prefer vanilla ice cream?
Answer:
0.17

Exercise 16.
Write a sentence explaining the meaning of the approximate relative frequency 0.55.
Answer:
Approximately 55% of students responding to the survey are males.

Exercise 17.
Write a sentence explaining the meaning of the approximate relative frequency 0.10.
Answer:
Approximately 10% of students responding to the survey are females who prefer chocolate ice cream.

Example 3.
In the previous exercises, you used the total number of students to calculate relative frequencies. These relative frequencies were the proportion of the whole group who answered the survey a certain way. Sometimes we use row or column totals to calculate relative frequencies. We call these row relative frequencies or column relative frequencies.
Below is the two-way frequency table for your reference. To calculate “the proportion of male students who prefer chocolate ice cream,” divide the 22 male students who preferred chocolate ice cream by the total of 45 male students. This proportion is \(\frac{22}{45}\) ≈ 0.49. Notice that you used the row total to make this calculation. This is a row relative frequency.

Exercises 18–22
In Exercise 13, you used the total number of students to calculate relative frequencies. These relative frequencies were the proportion of the whole group who answered the survey a certain way.

Exercise 18.
Suppose you are interested in the proportion of male students who prefer chocolate ice cream. How is this value different from “the proportion of students who are male and prefer chocolate ice cream”? Discuss this with your neighbor.
Answer:
The proportion of students who are male and prefer chocolate ice cream is \(\frac{22}{82}\) ≈ 0.27. This proportion uses all
82 students. The proportion of male students who prefer chocolate ice cream is \(\frac{22}{45}\) ≈ 0.49. This proportion uses only the 45 male students as its total.

Exercise 19.
Use the table provided in Example 3 to calculate the following relative frequencies.
a. What proportion of students who prefer vanilla ice cream are female?
\(\frac{14}{27}\) ≈ 0.52

b. What proportion of male students prefer strawberry ice cream? Write a sentence explaining the meaning of this proportion in the context of this problem.
Answer:
\(\frac{10}{45}\) ≈ 0.22 Twenty-two percent of male students in this survey prefer strawberry ice cream.

c. What proportion of female students prefer strawberry ice cream?
Answer:
\(\frac{15}{37}\) ≈ 0.41

d. What proportion of students who prefer strawberry ice cream are female?
Answer:
\(\frac{15}{25}\) ≈ 0.60

Exercise 20.
A student is selected at random from this school. What would you predict this student’s favorite ice cream to be? Explain why you chose this flavor.
Answer:
I would predict that the student’s favorite flavor is chocolate because more students chose chocolate in the survey.

Exercise 21.
Suppose the randomly selected student is male. What would you predict his favorite flavor of ice cream to be? Explain why you chose this flavor.
Answer:
I would predict his favorite flavor to be chocolate because more male students chose chocolate in the survey.

Exercise 22.
Suppose the randomly selected student is female. What would you predict her favorite flavor of ice cream to be? Explain why you chose this flavor.
Answer:
I would predict her favorite flavor to be strawberry because more female students chose strawberry in the survey.

Eureka Math Grade 8 Module 6 Lesson 13 Problem Set Answer Key

Every student at Abigail Douglas Middle School is enrolled in exactly one extracurricular activity. The school counselor recorded data on extracurricular activity and gender for all 254 eighth-grade students at the school.
The counselor’s findings for the 254 eighth-grade students are the following:
Of the 80 students enrolled in band, 42 are male.
Of the 65 students enrolled in choir, 20 are male.
Of the 88 students enrolled in sports, 30 are female.
Of the 21 students enrolled in art, 9 are female.

Question 1.
Complete the table below.

Answer:

Question 2.
Write a sentence explaining the meaning of the frequency 38 in this table.
Answer:
The frequency of 38 represents the number of eighth-grade students who are enrolled in band and are female.

Use the table provided above to calculate the following relative frequencies.
Question 3.
What proportion of students are male and enrolled in choir?
Answer:
\(\frac{20}{254}\) ≈ 0.08

Question 4.
What proportion of students are enrolled in a musical extracurricular activity (i.e., band or choir)?
Answer:
\(\frac{80+65}{254}\) ≈ 0.57

Question 5.
What proportion of male students are enrolled in sports?
Answer:
\(\frac{58}{132}\) ≈ 0.44

Question 6.
What proportion of students enrolled in sports are male?
Answer:
\(\frac{58}{88}\) ≈ 0.66

Pregnant women often undergo ultrasound tests to monitor their babies’ health. These tests can also be used to predict the gender of the babies, but these predictions are not always accurate. Data on the gender predicted by ultrasound and the actual gender of the baby for 1,000 babies are summarized in the two-way table below.

Question 7.
Write a sentence explaining the meaning of the frequency 130 in this table.
Answer:
The frequency of 130 represents the number of babies who were predicted to be female but were actually male (i.e., the ultrasound prediction was not correct for these babies).

Use the table provided above to calculate the following relative frequencies.
Question 8.
What is the proportion of babies who were predicted to be male but were actually female?
Answer:
\(\frac{48}{1000}\) = 0.048

Question 9.
What is the proportion of incorrect ultrasound gender predictions?
Answer:
\(\frac{130+48}{1000}\) = 0.178

Question 10.
For babies predicted to be female, what proportion of the predictions were correct?
Answer:
\(\frac{432}{562}\) ≈ 0.769

Question 11.
For babies predicted to be male, what proportion of the predictions were correct?
Answer:
\(\frac{390}{438}\) ≈ 0.890

Eureka Math Grade 8 Module 6 Lesson 13 Exit Ticket Answer Key

Question 1.
Explain what the term bivariate categorical data means.
Answer:
Bivariate categorical data means that the data set comprises data on two variables that are both categorical.

Question 2.
Explain how to calculate relative frequency. What is another word for relative frequency?
Answer:
Relative frequency is calculated by dividing a frequency by the total number of observations. Another word for relative frequency is proportion.

Question 3.
A random group of students are polled about how they get to school. The results are summarized in the table below.

Answer:

a. Calculate the relative frequencies for the table above. Write them as a percent in each cell of the table. Round to the nearest tenth of a percent.
Answer:
See the completed table above.

b. What is the relative frequency for the Carpool category? Write a sentence interpreting this value in the context of school transportation.
Answer:
The relative frequency is 0.333, or 33.3%. Approximately 33.3% of the students surveyed use a carpool to get to school.

c. What is the proportion of students who are female and walk to school? Write a sentence interpreting this value in the context of school transportation.
Answer:
The proportion is 0.069, or 6.9%. Approximately 6.9% of the students surveyed are female and walk to school.

d. A student is selected at random from this school. What would you predict this student’s mode of school transportation to be? Explain.
Answer:
I would predict the student would ride the bus because more students in the survey chose this mode of transportation.

Engage NY Eureka Math 8th Grade Module 6 Lesson 11 Answer Key

Eureka Math Grade 8 Module 6 Lesson 11 Exercise Answer Key

Exercise 1.
Old Faithful is a geyser in Yellowstone National Park. The following table offers some rough estimates of the length of an eruption (in minutes) and the amount of water (in gallons) in that eruption.

This data is consistent with actual eruption and summary statistics that can be found at the following links:
http://geysertimes.org/geyser.php?id = OldFaithful and http://www.yellowstonepark.com/2011/07/about – old – faithful/
a. Chang wants to predict the amount of water in an eruption based on the length of the eruption. What should he use as the dependent variable? Why?
Answer:
Since Chang wants to predict the amount of water in an eruption, the time length (in minutes) is the predictor, and the amount of water is the dependent variable.

b. Which of the following two scatter plots should Chang use to build his prediction model? Explain.

Answer:
The predicted variable goes on the vertical axis with the predictor on the horizontal axis. So, the amount of water goes on the y – axis. The plot on the graph on the right should be used.

c. Suppose that Chang believes the variables to be linearly related. Use the first and last data points in the table to create a linear prediction model.
Answer:
m = \(\frac{84000 – 3700}{4.5 – 1.5}\)≈1,566.7
So, y = a + (1,566.7)x.
Using either (1.5,3700) or (4.5,8400) allows students to solve for the intercept. For example, solving 3,700 = a + (1,566.7)(1.5) for a yields a = 1,349.95, or rounded to 1,350.0 gallons. Be sure students talk through the units in each step of the calculations.
The (informal) linear prediction model is y = 1,350.0 + 1,566.7x. The amount of water (y) is in gallons, and the length of the eruption (x) is in minutes.

d. A friend of Chang’s told him that Old Faithful produces about 3,000 gallons of water for every minute that it erupts. Does the linear model from part (c) support what Chang’s friend said? Explain.
Answer:
This question requires students to interpret slope. An additional minute in eruption length results in a prediction of an additional 1,566.7 gallons of water produced. So, Chang’s friend who claims Old Faithful produces 3,000 gallons of water a minute must be thinking of a different geyser.

e. Using the linear model from part (c), does it make sense to interpret the y – intercept in the context of this problem? Explain.
Answer:
No, it doesn’t make sense because if the length of an eruption is 0, then it cannot produce 1,350 gallons of water. (Convey to students that some linear models have y – intercepts that do not make sense within the context of a problem.)

Exercise 2.
The following table gives the times of the gold, silver, and bronze medal winners for the men’s 100 – meter race (in seconds) for the past 10 Olympic Games.

Data Source: https://en.wikipedia.org/wiki/100_metres_at_the_Olympics#Men

a. If you wanted to describe how mean times change over the years, which variable would you use as the independent variable, and which would you use as the dependent variable?
Answer:
Mean medal time (dependent variable) is being predicted based on year (independent variable).

b. Draw a scatter plot to determine if the relationship between mean time and year appears to be linear. Comment on any trend or pattern that you see in the scatter plot.
Answer:
The scatter plot indicates a negative trend, meaning that, in general, the mean race times have been decreasing over the years even though there is not a perfect linear pattern.

c. One reasonable line goes through the 1992 and 2004 data. Find the equation of that line.
Answer:
The slope of the line through (1992,10.01) and (2004,9.86) is \(\frac{10.01 – 9.86}{1992 – 2004}\) = – 0.0125.
To find the intercept using (1992,10.01), solve 10.01 = a + ( – 0.0125)(1992) for a, which yields
a = 34.91.
The equation that predicts the mean medal race time for an Olympic year is y = 34.91 + ( – 0.0125)x.
The mean medal race time (y) is in seconds, and the time (x) is in years.

Note to Teacher: In Algebra I, students learn a formal method called least squares for determining a “best – fitting” line. For comparison, the least squares prediction line is y = 34.3562 + ( – 0.0122)x.

d. Before he saw these data, Chang guessed that the mean time of the three Olympic medal winners decreased by about 0.05 second from one Olympic Game to the next. Does the prediction model you found in part (c) support his guess? Explain.
Answer:
The slope – 0.0125 means that from one calendar year to the next, the predicted mean race time for the top three medals decreases by 0.0125 second. So, between successive Olympic Games, which occur every four years, the predicted mean race time is reduced by 0.05 second because 4(0.0125) = 0.05.

e. If the trend continues, what mean race time would you predict for the gold, silver, and bronze medal winners in the 2016 Olympic Games? Explain how you got this prediction.
Answer:
If the linear pattern were to continue, the predicted mean time for the 2016 Olympics is 9.71 seconds because 34.91 – (0.0125)(2016) = 9.71.

f. The data point (1980,10.3) appears to have an unusually high value for the mean race time (10.3). Using your library or the Internet, see if you can find a possible explanation for why that might have happened.
Answer:
The mean race time in 1980 was an unusually high 10.3 seconds. In their research of the 1980 Olympic Games, students find that the United States and several other countries boycotted the games, which were held in Moscow. Perhaps the field of runners was not the typical Olympic quality as a result. Atypical points in a set of data are called outliers. They may influence the analysis of the data.

Eureka Math Grade 8 Module 6 Lesson 11 Problem Set Answer Key

Question 1.
From the United States Bureau of Census website, the population sizes (in millions of people) in the United States for census years 1790–2010 are as follows.

a. If you wanted to be able to predict population size in a given year, which variable would be the independent variable, and which would be the dependent variable?
Answer:
Population size (dependent variable) is being predicted based on year (independent variable).

b. Draw a scatter plot. Does the relationship between year and population size appear to be linear?
Answer:
The relationship between population size and year of birth is definitely nonlinear. Note that investigating nonlinear relationships is the topic of the next two lessons.

c. Consider the data only from 1950 to 2010. Does the relationship between year and population size for these years appear to be linear?
Answer:
Drawing a scatter plot using the 1950–2010 data indicates that the relationship between population size and year of birth is approximately linear, although some students may say that there is a very slight curvature to the data.

d. One line that could be used to model the relationship between year and population size for the data from 1950 to 2010 is y = – 4875.021 + 2.578x. Suppose that a sociologist believes that there will be negative consequences if population size in the United States increases by more than 2 \(\frac{3}{4}\) million people annually. Should she be concerned? Explain your reasoning.
Answer:
This problem is asking students to interpret the slope. Some students will no doubt say that the sociologist need not be concerned, since the slope of 2.578 million births per year is smaller than her threshold value of 2.75 million births per year. Other students may say that the sociologist should be concerned, since the difference between 2.578 and 2.75 is only 172,000 births per year.

e. Assuming that the linear pattern continues, use the line given in part (d) to predict the size of the population in the United States in the next census.
Answer:
The next census year is 2020.
– 4875.021 + (2.578)(2020) = 332.539
The given line predicts that the population then will be 332.539 million people.

Question 2.
In search of a topic for his science class project, Bill saw an interesting YouTube video in which dropping mint candies into bottles of a soda pop caused the soda pop to spurt immediately from the bottle. He wondered if the height of the spurt was linearly related to the number of mint candies that were used. He collected data using 1, 3, 5, and 10 mint candies. Then, he used two – liter bottles of a diet soda and measured the height of the spurt in centimeters. He tried each quantity of mint candies three times. His data are in the following table.

a. Identify which variable is the independent variable and which is the dependent variable.
Answer:
Height of spurt is the dependent variable, and number of mint candies is the independent variable because height of spurt is being predicted based on number of mint candies used.

b. Draw a scatter plot that could be used to determine whether the relationship between height of spurt and number of mint candies appears to be linear.
Answer:

c. Bill sees a slight curvature in the scatter plot, but he thinks that the relationship between the number of mint candies and the height of the spurt appears close enough to being linear, and he proceeds to draw a line.
Answer:
His eyeballed line goes through the mean of the three heights for three mint candies and the mean of the three heights for 10 candies. Bill calculates the equation of his eyeballed line to be
y = – 27.617 + (43.095)x,
where the height of the spurt (y) in centimeters is based on the number of mint candies (x). Do you agree with this calculation? He rounded all of his calculations to three decimal places. Show your work.
Yes, Bill’s equation is correct.
The slope of the line through (3,101.667) and (10,403.333) is \(\frac{403.333 – 101.667}{10 – 3}\) = 43.095.
The intercept could be found by solving 403.333 = a + (43.095)(10) for a, which yields
a = – 27.617.
So, a possible prediction line is y = – 27.617 + (43.095)x.

d. In the context of this problem, interpret in words the slope and intercept for Bill’s line. Does interpreting the intercept make sense in this context? Explain.
Answer:
The slope is 43.095, which means that for every mint candy dropped into the bottle of soda pop, the height of the spurt increases by 43.095 cm.
The y – intercept is (0, – 27.617). This means that if no mint candies are dropped into the bottle of soda pop, the height of the spurt is – 27.617 ft. This does not make sense within the context of the problem.

e. If the linear trend continues for greater numbers of mint candies, what do you predict the height of the spurt to be if 15 mint candies are used?
Answer:
– 27.617 + (43.095)(15) = 618.808
The predicted height would be 618.808 cm, which is slightly over 20 ft.

Eureka Math Grade 8 Module 6 Lesson 11 Exit Ticket Answer Key

Question 1.
According to the Bureau of Vital Statistics for the New York City Department of Health and Mental Hygiene, the life expectancy at birth (in years) for New York City babies is as follows.

Data Source: http://www.nyc.gov/html/om/pdf/2012/pr465 – 12_charts.pdf
a. If you are interested in predicting life expectancy for babies born in a given year, which variable is the independent variable, and which is the dependent variable?
Answer:
Year of birth is the independent variable, and life expectancy in years is the dependent variable.

b. Draw a scatter plot to determine if there appears to be a linear relationship between the year of birth and life expectancy.
Answer:
Life expectancy and year of birth appear to be linearly related.

c. Fit a line to the data. Show your work.
Answer:
Answers will vary. For example, the line through (2001,77.9) and (2009,80.6) is y = – 597.438 + (0.3375)x, where life expectancy (y) is in years, and the time (x) is in years.
Note to Teacher: The formal least squares line (Algebra I) is y = – 612.458 + (0.345)x.

d. Based on the context of the problem, interpret in words the intercept and slope of the line you found in part (c).
Answer:
Answers will vary based on part (c). The intercept says that babies born in New York City in Year 0 should expect to live around – 597 years! Be sure students actually say that this is an unrealistic result and that interpreting the intercept is meaningless in this problem. Regarding the slope, for an increase of 1 in the year of birth, predicted life expectancy increases by 0.3375 year, which is a little over four months.

e. Use your line to predict life expectancy for babies born in New York City in 2010.
Answer:
Answers will vary based on part (c).
– 597.438 + (0.3375)(2010) = 80.9
Using the line calculated in part (c), the predicted life expectancy for babies born in New York City in 2010 is 80.9 years, which is also the value given on the website.