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tankabanditka [31]
2 years ago
12

At home, Alice noticed that she had a flat tire. With the spare, Alice drove 30 miles to the next shop at 45 mph. After getting

the new tire, she drove home at 70 mph. What was Alice's average speed for her round trip?
Mathematics
1 answer:
Mrac [35]2 years ago
7 0

Answer: Approximately 54.7826 mph

As a fraction, this average speed is exactly 1260/23 mph.

========================================================

Work Shown:

"Alice drove 30 miles...at 45 mph". This means

t = d/r

t = 30/45

t = 2/3

She drove for 2/3 of an hour, which is (2/3)*60 = 40 minutes.

Then "she drove home at 70 mph". She drove the same distance, meaning,

t = d/r

t = 30/70

t = 3/7

She drove for 3/7 of an hour, which is roughly (3/7)*60 = 25.714 minutes.

------------------------------

Add up the time values

(2/3 hr) + (3/7 hr)

(2/3 + 3/7) hr

(14/21 + 9/21) hr

23/21 hr

She spent 23/21 of an hour driving going to the auto shop and then back home. The total distance she traveled was 30+30 = 60 miles.

Use these values to find the speed

r = d/t

r = d divided by t

r = 60 divided by (23/21)

r = (60/1) divided by (23/21)

r = (60/1) * (21/23)

r = (60*21)/(1*23)

r = 1260/23

r = 54.7826

Her average speed for the round trip is about 54.7826 mph.

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A local police chief claims that 31% of all drug-related arrests are never prosecuted. A sample of 500 arrests shows that 27% of
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Answer:

z=\frac{0.27 -0.31}{\sqrt{\frac{0.31(1-0.31)}{500}}}=-1.933  

p_v =P(Z  

If we compare the p value obtained and the significance level given \alpha=0.02 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 2% of significance the proportion of arrests that were not prosecuted is not significanlty less than 0.31.  

Step-by-step explanation:

1) Data given and notation

n=500 represent the random sample taken

X represent the number of arrests that were not prosecuted.

\hat p=0.27 estimated proportion of arrests that were not prosecuted

p_o=0.31 is the value that we want to test

\alpha=0.02 represent the significance level

Confidence=98% or 0.98

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.31.:  

Null hypothesis:p\geq 0.31  

Alternative hypothesis:p < 0.31  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.27 -0.31}{\sqrt{\frac{0.31(1-0.31)}{500}}}=-1.933  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.02. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

If we compare the p value obtained and the significance level given \alpha=0.02 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 2% of significance the proportion of arrests that were not prosecuted is not significanlty less than 0.31.  

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