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Arlecino [84]
3 years ago
9

the perimeter of this rectangle and area, and if the area was 100 what would the value of x be, also if the area is 200, what is

the value of x. thank you so much to whoever answers

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Given:

Length of the rectangle = x + 21

Width of the rectangle = 5

To find:

Perimeter and area of the rectangle

Solution:

Perimeter = 2(l + w)

                =2(x+21+5)

                =2(x+26)

Perimeter =2x+52 units

Area = length × width

        =(x+21)\times 5

Area =5x + 105 square units

Substitute perimeter is 100 and find the value of x:

100=2x+52

Subtract 52 from both sides.

100-52=2x+52-52

48=2x

Divide by 2 on both sides, we get

24=x

x = 24

Substitute area is 200 and find the value of x:

200=5x + 105

Subtract 105 from both sides.

200-105=5x + 105-105

95=5x

Divide by 5 on both sides, we get

19=x

x = 19

If the perimeter is 100, then the value of x is 24.

If the area is 200, then the value of x is 19.

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Kaylis [27]

Answer: 95% confidence interval = 20,000 ± 2.12\times\frac{1500}{\sqrt{17} }

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

Step-by-step explanation:

Given :

Sample size(n) = 17

Sample mean = 20000

Sample standard deviation = 1,500

5% confidence

∴ \frac{\alpha}{2} = 0.025

Degree of freedom (d_{f}) = n-1 = 16

∵ Critical value at ( 0.025 , 16 ) = 2.12

∴ 95% confidence interval = mean ± Z_{c}\times\frac{\sigma}{\sqrt{n} }

Critical value  at 95% confidence interval = 20,000 ± 2.12\times\frac{1500}{\sqrt{17} }

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

3 0
3 years ago
A counselor at a community college claims that the mean GPA for students who have transferred to State University is more than 3
Lena [83]

Answer:

t=\frac{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3  

If we compare the p value and the significance level for example \alpha=1-0.99=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

Step-by-step explanation:

Data given and notation  

\bar X=3.25 represent the sample mean

s=0.3 represent the sample standard deviation

n=36 sample size  

\mu_o =3.1 represent the value that we want to test  

\alpha 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 more than 3.1, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.1  

Alternative hypothesis:\mu > 3.1  

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{3.25-3.1}{\frac{0.3}{\sqrt{36}}}=3  

P-value  

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

df=n-1=36-1=35  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(35)}>3)=0.00247  

Conclusion  

If we compare the p value and the significance level for example \alpha=1-0.99=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis.

And the best conclusion would be:

a) Yes. The counselor's claim is valid

Because we reject the null hypothesis in favor to the alternative hypothesis.

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Answer:

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