Round 94.62 to the nearest square inch

The area of a rectangle is 3 3/4 unit squares, and its length is 5 units. Find the width and the perimeter of this rectangle.

dlinn [17]

Answer:

Width = 3/4 units or 0.75 units

perimeter of the rectangle = 23/2 = 11 1/2 units or 11.5 units

Step-by-step explanation:

A rectangle has two opposite sides equal in length. The 2 opposite sides are also parallel to each other.

The area of a rectangle = LW

where

L = length

W = width

The area of the rectangle = 3 3/4 units² = 15/4 units²

Length = 5 units

Therefore,

15/4 = 5W

multiply both sides by 4

15 = 20W

divide both sides by 20

W = 15/20

W = 3/4

Width = 3/4

Perimeter of the rectangle = 2L + 2W

Perimeter of the rectangle = 2(L + W)

Perimeter of the rectangle = 2(5 + 3/4)

perimeter of the rectangle = 2(23/4)

perimeter of the rectangle = 23/2 = 11 1/2 units or 11.5 units

6 0

3 years ago

?? Help me pls??? Thank you

Romashka [77]

it would be inconsistent. Meaning it won't have a solution

3 0

1 year ago

Alana took a total of 30 quizzes over the course of 5 weeks. How many weeks of school will Alana have to attend this quarter bef

vampirchik [111]

Step-by-step explanation:

We can create a ratio and solve.

$\frac{5}{30 } = \frac{x}{54}$

We then isolate for x and get 9

7 0

3 years ago

Read 2 more answers

From the list of these numbers 3,5,7,12,15,18,20<br> What is a factor of 10

il63 [147K]

The factors of 10 are 1, 2, 5, and 10. You can also look at this the other way around: if you can multiply two whole numbers to create a third number, those two numbers are factors of the third. 2 x 5 = 10, so 2 and 5 are factors of 10.

6 0

3 years ago

Use a t-distribution to find a confidence interval for the difference in means μd=μ1-μ2 using the relevant sample results from p

V125BC [204]

Answer:

a)

best estimate = Xd[bar]=4.80

margin of error = 8.66

The 99% confidence interval is -3.86 to 13.46

b)

test statistic = -1.86

p-value = 0.0526

Decision: Reject the null hypothesis.

At the 5% significance level, you can conclude that the population mean of the difference between treatment 1 and treatment 2 is less than zero.

Step-by-step explanation:

Hello!

a) 99% CI

Using d=X₁-X₂ to determine the study variable Xd: the difference between treatment 1 and treatment 2.

Assuming that this variable has an approximately normal distribution: Xd≈N(μd;σ²d)

To calculate the sample mean and standard deviation you have to calculate the difference between the values of both treatments first.

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5

22-18= 4 ; 27-29= -2 ; 32-25= 7; 26-20= 6 ; 29-20= 9

n= 5

Xd[bar]= ∑X/n= 24/5= 4.80

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/4*[186-(24²)/5]= 17.7

Sd= 4.21

The parameter of interes is the population mean od the difference, μd

The best estimate for this parameter is the sample mean, Xd[bar]=4.80

Using the t-distribution, the formula for the Confidence Interval is

Xd[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$

Where the margin of error is:

$t_{n-1;1-\alpha /2}$ * $\frac{Sd}{\sqrt{n} }$ = $t_{4;0.995}$ * $\frac{Sd}{\sqrt{n} }$ = 4.604* $\frac{4.21}{\sqrt{5} }$ = 8.66

99% CI [-3.86; 13.46]

b) 5% Hypothesis test

The variable of interest is defined d=X₁-X₂; Xd: the difference between treatment 1 and treatment 2. Xd≈N(μd;σ²d)

The statistic hypotheses are:

H₀: μd = 0

H₁: μd < 0

α: 0.05

The statistic to use for this test is:

$t_{H_0}= \frac{X_d[bar]-Mu_d}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

As before you have to calculate the difference between the observation for each case and then the sample mean and standard deviation:

Case 1 ; Case 2 ; Case 3 ; Case 4 ; Case 5 ; Case 6 ; Case 7 ; Case 8

18-18= 0; 12-19= -7; 11-25= -14; 21-21= 0; 15-19= -4; 11-14=-3; 14-15= -1; 22-20= 2

n= 8

Xd[bar]= ∑X/n= -27/8= -3.38

Sd²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[275-(-27²)/8]= 26.27

Sd= 5.13

$t_{H_0}= \frac{-3.38-0}{\frac{5.13}{\sqrt{8} } }= -1.86$

This test is one-tailed to the left, which means that you will reject the null hypothesis to small values of t, the p-value of the test has the same direction as the rejection region, this means that it is one-tailed to the left and you can calculate it as:

P(≤-1.86)= 0.0526

The decision rule using the p-value is:

If p-value > α, do not reject the null hypothesis.

If p-value ≤ α, reject the null hypothesis.

The p-value is greater than the significance level so the decision is to reject the null hypothesis.

I hope it helps!

7 0

4 years ago