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pav-90 [236]
3 years ago
7

the perimeter of a rectangle swimming pool is 486 feet. The width of the pool is 35 feet less than the length. Find the length a

nd the width​
Mathematics
2 answers:
statuscvo [17]3 years ago
5 0

Answer:

Length: 139 feet

Width: 104 feet

Step-by-step explanation:

The formula for the perimeter of a rectangle can be given by P = 2l + 2w. We are given the perimeter of the pool along with the width.

P = 486

w = l - 35

From here, all we have to do is plug back into the original formula:

486 = 2l + 2(l - 35)

Which can be further simplified as:

486 = 2l + 2l - 70

486 = 4l - 70

From here, all we have to do is add 70 to both sides of the equation and divide by four:

556 = 4l

139 = l

To make sure that this answer is accurate, we can find that the width of the rectangle should then be 104 (given by 139 - 35). All we have to do is plug back into the original equation:

P = 2l + 2w

P = 2(139) + 2(104)

P = 278 + 208

P = 486

And the substitution works, so the length of the rectangle would be 139 feet and the width would be 104 feet.

Delicious77 [7]3 years ago
3 0

Answer:

Length = 139 and Width = 104 .

Step-by-step explanation:

Given: The perimeter of a rectangle swimming pool is 486 feet. The width of the pool is 35 feet less than the length.

To find: Find the length and the width​ .

Solution: We have given that

Let the length = x

According to question

The width of the pool is 35 feet less than the length.

width = x-35

perimeter = 2(length + width)

plugging the values

486   = 2(x + x-35)

486  =  2x + 2x - 70

486  = 4x - 70

On adding by 70 both side

486 +70 = 4x-70 +70

556 = 4x

On dividing by 4 both side

139 = x

so, length = 139 .

    width = x -35 =  139- 35 = 104

Therefore,  length = 139 and width = 104.

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3 years ago
Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for the low income group,Test the claim that the m
miskamm [114]

Answer:

Null hypothesis:

\mathtt{H_o : \mu = 21.21}

Alternative hypothesis

\mathtt{H_1 : \mu \geq 21.21}

t = -0.080

Decision Rule: To  reject the null hypothesis if t > 1.340 at t

Since t = -0.080, this implies that t < 1.340 that means the t statistics value did not fall into the rejection region. Hence, we fail to reject the null hypothesis at the level of significance 0.10

Conclusion: We conclude that there is insufficient evidence to support the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm.

Step-by-step explanation:

Given that:

the sample mean \overline x = 21.15

the standard deviation \sigma = 4.7512

sample size N = 40

The objective is to test the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm.

At the level of significance of 0.1

The null hypothesis and the alternative hypothesis for this study can be computed as follows:

Null hypothesis:

\mathtt{H_o : \mu = 21.21}

Alternative hypothesis

\mathtt{H_1 : \mu \geq 21.21}

This test signifies a one-tailed test since the alternative is greater than or equal to 21.21

The t-test statistics can be computed by using the formula:

t= \dfrac{\overline x - \mu  }{\dfrac{\sigma}{\sqrt{n}}}

t = \dfrac{21.15- 21.21  }{\dfrac{4.7152}{\sqrt{40}}}

t = \dfrac{-0.06  }{\dfrac{4.7152}{6.3246}}

t = -0.080

degree of freedom = n - 1

degree of freedom =  40 - 1

degree of freedom = 39

From the t statistical tables,

at the level of significance ∝ = 0.1 and degree of freedom df = 39, the critical value of \mathtt{{T_{39,0.10} = 1.304}}

Decision Rule: To  reject the null hypothesis if t > 1.340 at t

Since t = -0.080, this implies that t < 1.340 that means the t statistics value did not fall into the rejection region. Hence, we fail to reject the null hypothesis at the level of significance 0.10

Conclusion: We conclude that there is insufficient evidence to support the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm.

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3 years ago
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