All categories

2 years ago

Find the volume of a rectangular prism with the dimensions:

Mathematics

2 answers:

larisa86 [58]2 years ago

8 0

Answer:

220

Step-by-step explanation:

V= whl = 5 · 11 · 4 = 220

Cerrena [4.2K]2 years ago

4 0

<h2><u>Question</u>:-</h2>

Find the volume of a rectangular prism with the dimensions:

length: 4 yd

width: 5 ft

height: 11 ft

<h3>Given:-</h3>

$\bullet$ Length of a rectangular prism = 4 yards = 4 × 3 = 12 feet.

$\bullet$ Breadth of a rectangular prism = 5 feet.

$\bullet$ Height of a rectangular prism = 11 feet.

The volume of a rectangular prism.

<h2>Solution:-</h2>

We know,

Formula of Volume of a rectangular prism is (Length × Breadth × Height) cubic units.

So, Volume = (12 × 5 × 11) cubic feet

Volume = 660 cubic feet

<h3>The volume of a rectangular prism is <u>6</u><u>6</u><u>0</u><u> </u><u>cubic </u><u>f</u><u>e</u><u>e</u><u>t</u>. [Answer]</h3>

You might be interested in

15. Which of the following statements is true? a. If the calculated value of F statistic is higher than the critical value, we r

quester [9]

Answer:

c) is true

Step-by-step explanation:

Given are four statements. Let us check one by one

a) If the calculated value of F statistic is higher than the critical value, we reject the alternative hypothesis in favor of the null hypothesis

False, we reject H0 in favour or alternate

b) The F statistic is always nonnegative as SSRr is never smaller than SSRur.

F statistics is non negative but reason is bot SSRr and SSRur are positive

Not because SSRr is never smaller than SSRur

c) Degrees of freedom of a restricted model is always less than the degrees of freedom of an unrestricted model.

This is true because in a restricted we put constraints thus df will be less

d) The F statistic is more flexible than the t statistic to test a hypothesis with a single restriction.

False. F statistic is for testing variances while t statistic for means.

4 0

3 years ago

A rectangular piece of paper has an area of 117 square inches and a perimeter of 44 inches. What are the dimensions of the paper

-Dominant- [34]

Answer:

13x9

Step-by-step explanation:

half of perimeret of rectangle L+W

half of perimeter of rectangle = 44/2 = 22

L+W=22

LxW=117

Think of 2 numbers that added =22 and multiplied =117

13+9 =22

13x9 =117

3 0

2 years ago

Find X please for geometry

inysia [295]

Answer:

x=22.5°

Step-by-step explanation:

∵it's a rhombus

∴each 2 opposite sides are parallel

∴the angel adjecent angle the angle 3x is equal to the angle 6x

∴ 3x+6x=180

8x=180

⇒x=22.5

3 0

3 years ago

Help me to do this question of math plz plz I will mark as brainlist plz plz plz plz

natulia [17]

Answer:

Alternate Angles are equal

8 0

3 years ago

A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn

Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, $\overline x_1$ = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, $\overline x_2$ = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

$\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

$t_{\alpha /2}$ = $t_{0.005, \, 21}$ = 1.721

Therefore, we have;

$\left (6.59- 15.5 \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}$

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0

2 years ago