Is this a function??????

A rectangular swimming pool measures 40 ft by 60 ft and is surrounded by a path of uniform width around the four edges. The peri

Alex_Xolod [135]

Answer:

6ft

Step-by-step explanation:

Lets x = the width of the path the surrounding path will add 2x to the pool dimension,therefore over all dimesion: (2x+40) by (2x+60) the overall perimeter (2x + 40) + 2(2x+60) = 248 Simplify divide by 2, result (2x+40) +(2x+60) = 124

 Combine like terms 2x + 2x +40 +60 =124 

4x + 100 = 124 

4x=124 = 100

4x=24

x=24/4

x= 6ft is the width of the path

check this by finding the perimeter with these values; 2x = 12 ft 

2 ( 12 + 40 ) + 2( 12 +60 ) 

2( 52 + 2(72)

104 + 144 = 248; confirms our solution of x= 6 ft

5 0

3 years ago

5 star + 20 pts + Branliest guarentee!!!

fiasKO [112]

Answer:

Figures A, C, and D. B isn't one since it is curved

7 0

3 years ago

Read 2 more answers

Mrs. Jameson asked all of her 110 students to list their birthdate and time on a sheet of paper and place it into a jar. One nam

tangare [24]

20/110 because there's 20 kids in her third period but 110 overall hope I helped in time

3 0

3 years ago

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2/3 cups of raisins are mixed with 1/2 cup peanuts what is the ratio

krok68 [10]

3:5 because if you add the fractions together it is what you get so the 5 would be raisins and the 3 would be peanuts

6 0

3 years ago

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of

Eva8 [605]

Answer:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

$\alpha=0.1$ 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 to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

Compute the test statistic

We don't know the population deviation, so for this case 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".

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

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

Conclusion

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

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

6 0

3 years ago