eureka math lesson 7 homework 2.8

Eureka Math Grade 5 Module 1 Lesson 7 Answer Key

Engage ny eureka math 5th grade module 1 lesson 7 answer key, eureka math grade 5 module 1 lesson 7 sprint answer key.

Question 1. 0 10 Answer:-  5

Question 2. 0 1 Answer:- 0.5

Question 3. 0 0.01 Answer:- 0.05

Question 4. 10 20 Answer:- 15

Question 5. 1 2 Answer:- 1.5

Question 6. 2 3 Answer:-2.5

Question 7. 3 4 Answer:- 3.5

Question 8. 7 8 Answer:- 7.5

Question 9. 1 2 Answer:- 1.5

Question 10. 0.1 0.2 Answer:- 0.15

Question 11. 0.2 0.3 Answer:- 0.25

Question 12. 0.3 0.4 Answer:- 0.35

Question 13. 0.7 0.8 Answer:- 0.75

Question 14. 0.1 0.2 Answer:- 0.5

Question 15. 0.01 0.02 Answer:- 0.015

Question 16. 0.02 0.03 Answer:- 0.025

Question 17. 0.03 0.04 Answer:- 0.035

Question 18. 0.07 0.08 Answer:- 0.075

Question 19. 6 7 Answer:- 6.5

Question 20. 16 17 Answer:- 16.5

Question 21. 38 39 Answer:- 38.5

Question 22. 0.4 0.5 Answer:- 0.45

Question 23. 8.5 8.6 Answer:- 8.55

Question 24. 2.8 2.9 Answer:- 2.85

Question 25. 0.03 0.04 Answer:- 0.035

Question 26. 0.13 0.14 Answer:- 0.135

Question 27. 0.37 0.38 Answer:- 0.375

Question 28. 80 90 Answer:- 85

Question 29. 90 100 Answer:- 95

Question 30. 8 9 Answer:- 8.5

Question 31. 9 10 Answer:- 9.5

Question 32. 0.8 0.9 Answer:- 0.85

Question 33. 0.9 1 Answer:- 0.95

Question 34. 0.08 0.09 Answer:- 0.085

Question 35. 0.09 0.1 Answer:- 0.095

Question 36. 26 27 Answer:- 26.5

Question 37. 7.8 7.9 Answer:- 7.85

Question 38. 1.26 1.27 Answer:- 1.265

Question 39. 29 30 Answer:- 29.5

Question 40. 9.9 10 Answer:- 9.95

Question 41. 7.9 8 Answer:- 7.95

Question 42. 1.59 1.6 Answer:- 1.595

Question 43. 1.79 1.8 Answer:- 1.795

Question 44. 3.99 4 Answer:- 3.995

Question 1. 10 20 Answer:- 15

Question 2. 1 2 Answer:- 1.5

Question 3. 0.1 0.2 Answer:- 0.5

Question 4. 0.01 0.02 Answer:- 0.015

Question 5. 0 10 Answer:- 5

Question 6. 0 1 Answer:- 0.5

Question 7. 1 2 Answer:- 1.5

Question 8. 2 3 Answer:- 2.5

Question 9. 6 7 Answer:- 6.5

Question 10. 1 2 Answer:- 1.5

Question 11. 0.1 0.2 Answer:- 0.15

Question 12. 0.2 0.3 Answer:- 0.25

Question 13. 0.3 0.4 Answer:- 0.35

Question 14. 0.6 0.7 Answer:- 0.65

Question 15. 0.1 0.2 Answer:- 0.15

Question 16. 0.01 0.02 Answer:- 0.015

Question 17. 0.02 0.03 Answer:- 0.025

Question 18. 0.03 0.04 Answer:- 0.035

Question 19. 0.06 0.07 Answer:- 0.065

Question 20. 7 8 Answer:- 7.5

Question 21. 17 18 Answer:- 17.5

Question 22. 47 48 Answer:- 47.5

Question 23. 0.7 0.8 Answer:- 0.75

Question 24. 4.7 4.8 Answer:- 4.75

Question 25. 2.3 2.4 Answer:- 2.35

Question 26. 0.02 0.03 Answer:- 0.025

Question 27. 0.12 0.13 Answer:- 0.125

Question 28. 0.47 0.48 Answer:- 0.475

Question 29. 80 90 Answer:- 85

Question 30. 90 100 Answer:- 95

Question 31. 8 9 Answer:- 8.5

Question 32. 9 10 Answer:- 9.5

Question 33. 0.8 0.9 Answer:- 0.85

Question 34. 0.9 1 Answer:- 0.95

Question 35. 0.08 0.09 Answer:- 0.085

Question 36. 0.09 0.1 Answer:- 0.095

Question 37. 36 37 Answer:- 36.5

Question 38. 6.8 6.9 Answer:- 6.85

Question 39. 1.46 1.47 Answer:- 1.465

Question 40. 39 40 Answer:- 39.5

Question 41. 9.9 10 Answer:- 9.95

Question 42. 6.9 7 Answer:- 6.95

Question 43. 1.29 1.3 Answer:- 1.295

Question 44. 6.99 7 Answer:- 6.995

Eureka Math Grade 5 Module 1 Lesson 7 Problem Set Answer Key

Fill in the table, and then round to the given place. Label the number lines to show your work. Circle the rounded number.

Eureka Math Grade 5 Module 1 Lesson 7 Exit Ticket Answer Key

Use the table to round the number to the given places. Label the number lines, and circle the rounded value. 8.546

Eureka Math Grade 5 Module 1 Lesson 7 Homework Answer Key

Question 4. On a Major League Baseball diamond, the distance from the pitcher’s mound to home plate is 18.386 meters. a. Round this number to the nearest hundredth of a meter. Use a number line to show your work.

b. How many centimeters is it from the pitcher’s mound to home plate?

Question 5. Jules reads that 1 pint is equivalent to 0.473 liters. He asks his teacher how many liters there are in a pint. His teacher responds that there are about 0.47 liters in a pint. He asks his parents, and they say there are about 0.5 liters in a pint. Jules says they are both correct. How can that be true? Explain your answer.

Leave a Reply Cancel reply

You must be logged in to post a comment.

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Eureka Math Grade 7 Module 4 Lesson 8 Answer Key

Engage ny eureka math 7th grade module 4 lesson 8 answer key, eureka math grade 7 module 4 lesson 8 example answer key.

Example 1: How Far Off? Find the absolute error for the following problems. Explain what the absolute error means in context. a. Taylor’s Measurement 1 Answer: |15 \(\frac{2}{8}\)in. – 15 in.| = |0.25 in| = 0.25 in. Taylor’s Measurement 1 was 0.25 in. away from the actual value of 15 in.

b. Connor’s Measurement 1 Answer: |15\(\frac{4}{8}\)in. – 15 in.| = |0.5 in.| = 0.5 in. Connor’s Measurement 1 was 0.5 in. away from the actual value of 15 in.

c. Jordan’s Measurement 2 Answer: |14\(\frac{6}{8}\) in. – 15 in.| = 0.25 in. Jordan’s Measurement 2 was 0.25 in. away from the actual value of 15 in.

Example 2: How Right Is Wrong? a. Find the percent error for Taylor’s Measurement 1. What does this mean? Answer: \(\frac{\left|15 \frac{2}{8} – 15\right|}{|15|}\) × 100% \(\frac{|0.25|}{|15|}\) × 100% \(\frac{1}{60}\) × 100% 1\(\frac{2}{3}\)% This means that Taylor’s measurement of 15.25 in. has an error that is 1\(\frac{2}{3}\)% of the actual value.

b. From Example 1, part (b), find the percent error for Connor’s Measurement 1. What does this mean? Answer: \(\frac{0.5 \mathrm{in} .}{15 \mathrm{in} .}\) × 100% = 3\(\frac{1}{3}\)% This means that Connor’s measurement of 15\(\frac{4}{8}\)in. has an error that is 3\(\frac{1}{3}\)% of the actual value.

c. From Example 1, part (c), find the percent error for Jordan’s Measurement 2. What does it mean? Answer: \(\frac{0.25 \mathrm{in} .}{15 \mathrm{in} .}\) × 100% = 1\(\frac{2}{3}\)% This means that Jordan’s measurement of 14\(\frac{6}{8}\)in. has an error that is 1\(\frac{2}{3}\)% of the actual value.

d. What is the purpose of finding percent error? Answer: It tells you how big your error is compared to the true value. An error of 1 cm is very small when measuring the distance for a marathon, but an error of 1 cm is very large if you are a heart surgeon. In evaluating the seriousness of an error, we usually compare it to the exact value.

Example 3: Estimating Percent Error The attendance at a musical event was counted several times. All counts were between 573 and 589. If the actual attendance number is between 573 and 589, inclusive, what is the most the percent error could be? Explain your answer. Answer: The most the absolute error could be is |589 – 573| = 16. The percent error will be largest when the exact value is smallest. Therefore, the most the percent error could be is \(\frac{16}{573}\) × 100%<2.8%. In this case, the percent error is less than 2.8%.

Eureka Math Grade 7 Module 4 Lesson 8 Exercise Answer Key

Calculate the percent error for Problems 1–3. Leave your final answer in fraction form, if necessary. Exercise 1. A real estate agent expected 18 people to show up for an open house, but 25 attended. Answer: \(\frac{|18 – 25|}{|25|}\) × 100% = 28%

Exercise 2. In science class, Mrs. Moore’s students were directed to weigh a 300 – gram mass on the balance scale. Tina weighed the object and reported 328 grams. Answer: \(\frac{|328 – 300|}{|300|}\) × 100% = 9\(\frac{1}{3}\)%

Exercise 3. Darwin’s coach recorded that he had bowled 250 points out of 300 in a bowling tournament. However, the official scoreboard showed that Darwin actually bowled 225 points out of 300. Answer: \(\frac{|250 – 225|}{|225|}\) × 100% = 11 \(\frac{1}{9}\)%

Eureka Math Grade 7 Module 4 Lesson 8 Problem Set Answer Key

Question 1. The odometer in Mr. Washington’s car does not work correctly. The odometer recorded 13.2 miles for his last trip to the hardware store, but he knows the distance traveled is 15 miles. What is the percent error? Use a calculator and the percent error formula to help find the answer. Show your steps. Answer: 15 is the exact value, and 13.2 is the approximate value. Using the percent error formula, \(\frac{|a – x|}{|x|}\) × 100%, the percent error is \(\frac{|13.2 – 15|}{|15|}\) × 100% = 12%. The percent error is equal to 12%.

b. Find the percent error of the area of the soccer field. Answer: Actual area: A = l × w A = (500)(250) = 125,000 The actual area is 125,000 square feet.

Approximate area: A = l × w A = (493)(246.5) The approximate area is 121,524.5 square feet.

Percent error of the area: \(\frac{|121,524.5 – 125,000|}{|125,000|}\) × 100% = 2.8%

c. Explain why the values from parts (a) and (b) are different. Answer: In part (a), 1.4% is the percent error for the length, which is one dimension of area. Part (b) is the percent error for the area, which includes two dimensions―length and width. The percent error for the width of the soccer field should be the same as the percent error for the length if the same measuring tool is used. So, 2.8% = 1.4% × 2. However, this is not always the case. Percent error for the width is not always the same as the percent error for the length. It is possible to have an error for both the length and the width, yet the area has no error. For example: publicized length = 100 feet, publicized width = 90 feet, actual length = 150 feet, and actual width = 60 feet.

Question 3. Kayla’s class went on a field trip to an aquarium. One tank had 30 clown fish. She miscounted the total number of clown fish in the tank and recorded it as 24 fish. What is Kayla’s percent error? Answer: \(\frac{|24 – 30|}{|30|}\) × 100% = 20%

b. Find the percent error in Sid’s estimation to the nearest hundredth using the π key on your calculator. \(\frac{|52 – 16 \pi|}{|16 \pi|}\) × 100% ≈ 3.45%

Question 5. The exact value for the density of aluminum is 2.699 g/cm 3 . Working in the science lab at school, Joseph finds the density of a piece of aluminum to be 2.75 g/cm 3 . What is Joseph’s percent error? (Round to the nearest hundredth.) Answer: \(\frac{|2.75 – 2.699|}{|2.699|}\) × 100% ≈ 1.89%

Question 6. The world’s largest marathon, The New York City Marathon, is held on the first Sunday in November each year. Between 2 million and 2.5 million spectators will line the streets to cheer on the marathon runners. At most, what is the percent error? Answer: \(\frac{|2.5 – 2|}{|2|}\) × 100% = 25%

b. Find the percent error of Jared’s estimate to two decimal places using the π key on your calculator. Answer: \(\frac{\left|127 – \pi 6 \cdot 3^{2}\right|}{\left|\pi 6 \cdot 3^{2}\right|}\) × 100% ≈ 1.85%. The percent error is approximately 1.85%.

c. Do you think Jared’s estimate was reasonable? Answer: Yes. The percent error is less than 2%.

d. Would this method of computing the area of a circle always be too large? Answer: Yes. If the circle has radius r, then the area of the circle is πr 2 , and the area of the square is 4r 2 . \(\frac{\pi r^{2}}{4 r^{2}}\) = \(\frac{\pi}{4}\). The area approximately equals 0.785 = 78.5%<80%.

Question 8. In a school library, 52% of the books are paperback. If there are 2,658 books in the library, how many of them are not paperback to the nearest whole number? Answer: 100% – 52% = 48% Let n represent the number of books that are not paperback. n = 0.48(2,658) n = 1,275.84 About 1,276 books are not paperback.

Question 9. Shaniqua has 25% less money than her older sister Jennifer. If Shaniqua has $180, how much money does Jennifer have? Answer: 100% – 25% = 75% Let j represent the amount of money that Jennifer has. 180 = \(\frac{3}{4}\)j \(\frac{4}{3}\)(180) = (\(\frac{3}{4}\))(\(\frac{4}{3}\))j 240 = j Jennifer has $240.

Question 10. An item that was selling for $1,102 is reduced to $806. To the nearest whole, what is the percent decrease? Answer: Let p represent the percent decrease. 1,102 – 806 = 296 296 = p ∙ 1,102 \(\frac{296}{1,10.2}\) = p 0.2686 = p The percent decrease is approximately 27%.

Question 11. If 60 calories from fat is 75% of the total number of calories in a bag of chips, find the total number of calories in the bag of chips. Answer: Let t represent the total number of calories in a bag of chips. 60 = \(\frac{3}{4}\) t \(\frac{4}{3}\) ∙ 60 = \(\frac{3}{4}\) ∙ \(\frac{4}{3}\) ∙ t 80 = t The total number of calories in the bag of chips is 80 calories.

Eureka Math Grade 7 Module 4 Lesson 8 Exit Ticket Answer Key

Question 1. The veterinarian weighed Oliver’s new puppy, Boaz, on a defective scale. He weighed 36 pounds. However, Boaz weighs exactly 34.5 pounds. What is the percent of error in measurement of the defective scale to the nearest tenth? Answer: \(\frac{|36 – 34.5|}{|34.5|}\) × 100% = 4 \(\frac{8}{23}\)% ≈ 4.3%

Question 2. Use the π key on a scientific or graphing calculator to compute the percent of error of the approximation of pi, 3.14, to the value π. Show your steps, and round your answer to the nearest hundredth of a percent. Answer: \(\frac{|3 \cdot 14 – \pi|}{|\pi|}\) × 100% = 0.05%

Question 3. Connor and Angie helped take attendance during their school’s practice fire drill. If the actual count was between 77 and 89, inclusive, what is the most the absolute error could be? What is the most the percent error could be? Round your answer to the nearest tenth of a percent. Answer: The most the absolute error could be is |89 – 77| = |12| = 12. The percent error will be largest when the exact value is smallest. The most the percent error could be is \(\frac{|12|}{|77|}\) × 100% < 15.6%. The percent error is less than 15.6%.

Leave a Comment Cancel Reply

You must be logged in to post a comment.