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3 years ago

What is the width of a rectangular prism with a length of 13 feet, volume of 11232 cubic feet, and height of 36 feet?

Mathematics

1 answer:

Fynjy0 [20]3 years ago

6 0

Answer:

24 ft.

Step-by-step explanation:

Volume (V) equals the product of length (l), width (w), and height (h)

V = l × w × h

Plug in the measurement you have into the equation

11232 = 13 × w × 36

Multiply 13 and 36

11232 = 468 × w

Divide both sides by 468

24 = w

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Write an equation of the direct variation that includes the point (5,1)

V125BC [204]

Answer:

y = 5x

Step-by-step explanation:

X=5 and Y=1

Direct variation can be represented by the equation y=kx.

Therefore,

1 = k(5)

5 = k

Now you use, k = 5 to write the direct variation equation.

Which gives you the answer y = 5x

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3 years ago

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What is the measure for the following angle?

Sonbull [250]

Answer:

The answer is 80 degrees, I believe

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3 years ago

Simplify. 3 – [–(–1)] <br> A. 4 <br> B. –2 <br> C. 2 <br> D. –4

Nesterboy [21]

Answer:

Step-by-step explanation:

3-(+1)

3-1

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3 years ago

What is a system of equations?

zlopas [31]

Answer:

A system of equations is when two or more equations have the same variable(s) that you work with together.

Step-by-step explanation:

For these types of equations, we typically use substitution, elimination, and graphing.

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3 years ago

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

3 years ago