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nika2105 [10]
3 years ago
5

The distance between two schools,A and B is 2 km.A market is situated a third of the distance from A to B.How far is the market

from B?
Mathematics
1 answer:
bekas [8.4K]3 years ago
3 0

Answer:

\frac{4}{3}

Step-by-step explanation:

Find distance between B and market: |B-M|=|M-B|

Distance between A and B: |A-B| = 2 , call this equation (1)

Distance between A and market |A-M| = \frac{1}{3} |A-B|=\frac{2}{3}, call this equation (2)

Solve (1) - (2)

|A-B-A+M| = 2-\frac{2}{3} \\|M-B|=|B-M|=\frac{4}{3}

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Answer:

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

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

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Step-by-step explanation:

Data given and notation  

\bar X=5.25 represent the sample mean

s=3.98 represent the sample standard deviation

n=54 sample size  

\mu_o =4.73 represent the value that we want to test  

\alpha 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 mean differs from 4.73, the system of hypothesis would be:  

Null hypothesis:\mu =4.73  

Alternative hypothesis:\mu \neq 4.73  

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-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:  

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

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

p_v =2*P(t_{53}>0.9601)=0.3414  

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