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solong [7]
3 years ago
8

(CO 5) An advocacy group claims that the mean braking distance of a certain type of tire is 75 feet when the car is going 40 mil

es per hour. In a test of 45 of these tires, the braking distance has a mean of 78 and a population standard deviation of 5.9 feet. Find the standardized test statistic and the corresponding p-value. Group of answer choices
Mathematics
1 answer:
KIM [24]3 years ago
4 0

Answer:

z=0.5084

p=0.2809

Step-by-step explanation:

Since number of tires tested for breaking distance is enough (>30), the standardized test statistic is calculated by the formula:

z=\frac{X-M}{\frac{s}{\sqrt{N} }} } where

  • X is the mean breaking distance in the sample (78)
  • M is the mean breaking distance  (75)
  • s is the standard deviation (5.9)
  • N is the sample size

Putting the numbers in the formula:

z=\frac{78-75}{\frac{5.9}{\sqrt{N} }} } =0.5084 and corresponding p value is:

p=0.28

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b) amount in the bank after 7 years if interest is compounded quarterly is $6,612.57

Step-by-step explanation:

We are given:

Principal Amount P= 5000

Rate r= 4% = 0.04

time t = 7 years

The formula used is: A=P(1+\frac{r}{n})^{nt}

where A is future value, P is principal amount, r is rate, n is compounded value and t is time

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Using values given in question and finding A

A=P(1+\frac{r}{n})^{nt}\\A=5000(1+\frac{0.04}{4})^{4*7} \\A=5000(1+0.01)^{28}\\A=5000(1.01)^{28}\\A=5000(1.321)\\A=6,605

So, amount in the bank after 7 years if interest is compounded quarterly is $6,605

b) Find the amount in the bank after 7 years if interest is compounded monthly?

If interest is compounded quarterly then n = 12

Using values given in question and finding A

A=P(1+\frac{r}{n})^{nt}\\A=5000(1+\frac{0.04}{12})^{12*7} \\A=5000(1+0.003)^{84}\\A=5000(1.003)^{84}\\A=5000(1.322)\\A=6,612.57

So, amount in the bank after 7 years if interest is compounded quarterly is $6,612.57

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