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KonstantinChe [14]
3 years ago
9

yler and Bethany have 240 feet of fencing with which to build a garden. Tyler wants the garden to be in the shape of a square, w

hile Bethany wants it to be rectangular with a length of 50 feet and a width of 70 feet. Which design would give the maximum area for the garden? Explain. Bethany’s design would give the larger garden because the area would be 3,500 ft2. Tyler’s design would give the larger garden because the area would be 3,600 ft2. Bethany’s and Tyler’s designs would give a garden with the same area. Bethany’s and Tyler’s designs are not possible with the given amount of fencing.
Mathematics
1 answer:
Sidana [21]3 years ago
4 0
To get which design would have maximum area we need to evaluate the area for Tyler's design. Given that the design is square, let the length= xft,  width=(120-x)
thus:
area will be:
P(x)=x(120-x)
P(x)=120x-x²
For maximum area P'(x)=0
P'(x)=120-2x=0
thus
x=60 ft
thus for maximum area x=60 ft
thus the area will be:
Area=60×60=3600 ft²
Thus we conclude that Tyler's design is the largest. Thus:
 the answer is:
<span>Tyler’s design would give the larger garden because the area would be 3,600 ft2. </span>
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Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

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

\alpha=0. 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 mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but 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{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

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

df=n-1=1700-1=1699  

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

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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