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andreev551 [17]

3 years ago

7

What is the surface area of a cube in which each face of the cube has an area of 7 cm^2?

1 answer:

prisoha [69]3 years ago

4 0

Answer:

In a cube, the length and width are the same, so both sides will be 7 cm, yielding an area of 49 square centimeters. The surface area of a 7 cm cube will be 294 cm2 . Hope this helps!

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What’s the answer I forgot half of this stuff‍♀️

earnstyle [38]

Answer:

your answer is 12/7

=1.714286

4 0

2 years ago

The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

$\frac{X-\mu}{\sigma} = Z$

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0

3 years ago

ken wants to rent a car for a week and to pay to more than $130. how far can he drive if the car rental cost $94 a week plus $0.

cupoosta [38]

If Ken wants to pay more than 130, he needs to drive more than 90 miles. You subtract 94 from 130, which equals 36. 36 divided by .4 (for the price per mile) and it equals 90.

4 0

3 years ago

Pls help with this and show work on piece of paper

Lapatulllka [165]

Answer:

????

i dont know

5 0

3 years ago

Read 2 more answers

Complete the assignment on a separate sheet of paper<br><br> Please attach pictures of your work.

Irina18 [472]

Answer:

<u>TO FIND :-</u>

Length of all missing sides.

<u>FORMULAES TO KNOW BEFORE SOLVING :-</u>

$\sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}$
$\cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}$
$\tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}$

<u>SOLUTION :-</u>

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

$=> \sin 16 = \frac{7}{x}$

$=> \frac{7}{x} = 0.27563......$

$=> x = \frac{7}{0.27563....} = 25.39568.....$ ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

$=> \sin 29 = \frac{6}{x}$

$=> \frac{6}{x} = 0.48480......$

$=> x = \frac{6}{0.48480....} = 12.37599.....$ ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

$=> \sin 30 = \frac{x}{11}$

$=> \frac{x}{11} = 0.5$

$=> x = 0.5 \times 11 = 5.5$

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

$=> \cos 43 = \frac{x}{12}$

$=> \frac{x}{12} = 0.73135......$

$=> x = 12 \times 0.73135.... = 8.77624....$ ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

$=> \cos 55 = \frac{x}{6}$

$=> \frac{x}{6} = 0.57357......$

$=> x = 6 \times 0.57357.... = 3.44145....$ ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

$=> \cos 73 = \frac{8}{x}$

$=> \frac{8}{x} = 0.29237......$

$=> x = \frac{8}{0.29237.....} = 27.36242.....$ ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

$=> \tan 69 = \frac{12}{x}$

$=> \frac{12}{x} = 2.60508......$

$=> x = \frac{12}{2.60508....} = 4.60636....$ ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

$=> \tan 20 = \frac{11}{x}$

$=> \frac{11}{x} = 0.36397......$

$=> x = \frac{11}{0.36397....} =30.22225....$ ≈ 30.2

5 0

2 years ago