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pochemuha
3 years ago
14

A grocery, a cafe, a boutique, and a dance studio have the addresses 500 Elm, 524 Elm, 538 Elm, and 544 Elm, but not necessarily

in that order. The boutique is between the cafe and the studio. The studio has the highest address. The grocery is between the cafe and boutique.
Mathematics
2 answers:
Alex3 years ago
5 0

Answer:

B

Step-by-step explanation:

On edge

rewona [7]3 years ago
3 0

Answer:

500 Elm: Cafe

524 Elm: Grocery

538 Elm: Boutique

544 Elm: Dance studio

Step-by-step explanation:

We know that the dance studio has the highest address, so 544.

For the others let's examine the possibilities.

The boutique is between the cafe and the studio, that means the following are possible:

Boutique can be at either 524 or 538, cannot be at 500 (since cafe has to be on the its left.

Cafe can be 500 or 524, cannot be 538 since boutique has to be between cafe and studio.

<u>The grocery is between the cafe and boutique. </u>Meaning the grocery cannot be 500 nor 538, since it has to be in the middle of the 3 remaining businesses.  So, grocery is at 524.

Meaning the Cage is at 500 and the boutique is at 538.

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Another way to write the expression the answer is this (t.14)-(t.5)
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Answer:

dy/dx=8

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3 years ago
The American Water Works Association reports that the per capita water use in a single-family home is 63 gallons per day. Legacy
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Answer:

t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

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p_v =P(t_{(27)}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=60 represent the sample mean

s=8.9 represent the sample standard deviation

n=28 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t 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 less than 63, the system of hypothesis would be:  

Null hypothesis:\mu \geq 63  

Alternative hypothesis:\mu < 63  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=28-1=27  

Since is a one side lower test the p value would be:  

p_v =P(t_{(27)}  

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If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

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