open to correct answers Q#1: Before beginning a fundraiser, Lindsey estimated that the fundraiser would earn $500. She sold 114 raffle tickets at $5 each for the fundraiser, actually earning $5x114=$570 for the fundraiser. What was the percent error in Lindsey’s estimate? Round to the nearest tenth of a percent, if necessary. A. 12.3% B. 14% C. 70% D. 77.2% my answer : C.70% fan and medal
the change from 500 to 570 is 70 (570 - 500 = 70) Divide this difference by the original amount: 70/500 = ???
0.14
which is 14%
so my correct answer is B
yes
Julie counted 25 players on a visiting soccer team. The actual number of players on the visiting soccer team was 27. What is the percent error in Julie’s measurement? If necessary, round to the nearest percent. my answer : 15%
I made a mistake and wasn't thinking. The actual answer to the first one is 12.3% and here's why https://www.mathsisfun.com/numbers/percentage-error.html using the formula on that page, we will get |50-57|/57 = 7/57 = 0.122807 which turns into 12.3% when you round for the other problem, it would be |25-27|/27 = 2/27 = 0.074 = 7.4% which rounds to 7%
okay
for this problem the answer is 7% and not 15%
yes for the second one
A car mechanic tested the accuracy of three speedometers. The table shows the speed registered by each of the three speedometers. The actual speed each speedometer was tested on was 90 kmph. Which statements are true? Speedometer Speed displayed (kmph) A 92 B 87 C 88 Choose exactly two answers that are correct. A. Speedometer A and Speedometer C had the same percent error. B. Speedometer C had the least percent error. C. Speedometer C had a 0.02% error. D. Speedometer B had about a 3.3% error. my answers : A & B
A and B are contradictions. How can you have A and C being the same, but then have C be smaller?
u can't C is way smaller than A which is bigger put those two together it doesn't make sense
i think it's A & D it's a 50-50 percent chance
A) Speedometer A and Speedometer C had the same percent error. A is true because both are roughly 2.2%
D) Speedometer B had about a 3.3% error. that's true because the error for B is |87-90|/90 = 3/90 = 1/30 = 0.033 = 3.3%
so i'm right?
yes it's A & D
David measured the distance from his house to a soccer field as 3 km. The actual distance from his house to the soccer field is 2.84 km. What is the approximate percent error in David’s measurement? Round to the nearest tenth of a percent, if necessary. A. 0.16% B. 5.3% C. 5.6% D. 16% my answer : C 5.6%
I agree
Akira estimated that a tree was 40 years old. The true age of the tree was 47 years old. Which expression can be used to calculate the percent error in Akira’s estimate? A.47-40/40x100% B.40-47/47x100% C.[40-47]/40x100% D.[ 40-47]/47x100% my answer : B
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