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Lunna [17]
3 years ago
11

Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim

e? mh
Mathematics
1 answer:
vivado [14]3 years ago
5 0

The average speed of Joshua during that time is 2500 m/h.

Explanation:

It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.

The total time taken by Joshua from 5:15 pm to 8:09 pm is

2 { h } 54 \text { min }=2 \mathrm{h}+\frac{54}{60} {h}

Dividing we get,

2 { h } 54 \text { min }=2 \mathrm{h}+0.9 {h}

Adding, we have,

2 { h } 54 \text { min }=2.9 {h}

Thus, the total time taken by Joshua is 2.9 {h}

To determine the average speed we use the formula,

speed=\frac{distance}{time}

where distance=7250 and time=2.9

Hence, substituting the values we have,

speed=\frac{7250}{2.9}

Dividing, we get,

speed=2500

Thus, the average speed of Joshua during that time is 2500 m/h.

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p_v =2*P(z>3.162)=0.0016    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Step-by-step explanation:

Data given and notation    

\bar X=1.02 represent the sample mean

\sigma=0.04 represent the population standard deviation    

n=40 sample size    

\mu_o =1 represent the value that we want to test  

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

z would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the true mean is equal to 1 pound or no :    

Null hypothesis:\mu =1    

Alternative hypothesis:\mu \neq 1    

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

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

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

P-value    

Since is a two-sided test the p value would be:    

p_v =2*P(z>3.162)=0.0016    

Conclusion    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.    

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3 years ago
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