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3 years ago

A rectangular prism have the dimensions, 3.5ft by 4ft by 5.2 ft. What is the volume of this 2

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

Answer:

Step-by-step explanation:

The units are cubic feet or ft^3

V = L * W * H

L = 5.2

W = 4

H = 3.5

All dimensions are in feet.

V = 5.2 * 4 * 3.5

V = 72.8 ft^3

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Bonus: Mr. Rao is planning on making a shed in his backyard (see his plan in the image below). However, local laws require sheds

Leviafan [203]

Answer:

North 10 feet, and East 5 feet.

Step-by-step explanation:

So we know that North is ^ way. We also know that East is > way.

These are the two directions we must go, since the property line is both South and West of the shed.

The shed is 10ft East of the property line, and we must be 15ft in that direction. So we must go another 5ft East.

Now, the shed is 5ft North of the property line, and we must be 15ft in that direction. So we must go another 10 feet North.

Hope this helps!

6 0

3 years ago

Plss help me as soon as possible i will give you lots of points

mezya [45]

Answer:

Step-by-step explanation:

75x + -300 = 25x + 200

Reorder the terms:

-300 + 75x = 25x + 200

Reorder the terms:

-300 + 75x = 200 + 25x

Solving

-300 + 75x = 200 + 25x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-25x' to each side of the equation.

-300 + 75x + -25x = 200 + 25x + -25x

Combine like terms: 75x + -25x = 50x

-300 + 50x = 200 + 25x + -25x

Combine like terms: 25x + -25x = 0

-300 + 50x = 200 + 0

-300 + 50x = 200

Add '300' to each side of the equation.

-300 + 300 + 50x = 200 + 300

Combine like terms: -300 + 300 = 0

0 + 50x = 200 + 300

50x = 200 + 300

Combine like terms: 200 + 300 = 500

50x = 500

Divide each side by '50'.

x = 10

Simplifying

x = 10

3 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$