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Sergeu [11.5K]
3 years ago
6

Becky runs 3 miles in 24 minutes. At the same rate, how many miles would she run in 60 minutes?

Mathematics
2 answers:
Scrat [10]3 years ago
7 0

Answer:

7.5 miles

Step-by-step explanation:

Speed = Distance / Time

Speed = \frac{3}{24} = \frac{1}{8} miles per minute

\frac{1}{8} * 60 = 7.5 miles

jeka943 years ago
7 0

Answer:

7.5 miles

Step-by-step explanation:

In order to answer this question we need to find how many minutes it takes Becky to run one mile.

To do this we need to divide the 24 minutes by the 3miles.

24/3 = 8

Therefore, it takes Becky 8 minutes per mile.

Next, to find how many miles she runs in an hour we need to divide 60 by 8.

60/8 = 7.5

Becky will run 7.5 miles per hour.

<em>I hope this helps!</em>

<em>- Kay :)</em>

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3 years ago
An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t
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Answer:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

\bar X=46.5 represent the mean

\sigma=1.1 represent the population standard deviation

n=150 sample size  

\mu_o =46.7 represent the value to verify

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:  

Null hypothesis:\mu = 46.7  

Alternative hypothesis:\mu \neq 46.7  

Since we know the population deviation the statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing we got:

z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23    

The p value would be given by:

p_v =2*P(z  

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

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