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insens350 [35]
3 years ago
15

A plastic box has the dimensions shown. 2 square pyramids are stacked on their bases. The bases are 8 centimeters by 8 centimete

rs and the height of each pyramid is 7.5 centimeters. What is the volume of the box? 160 cm3 320 cm3 480 cm3 960 cm3
Mathematics
2 answers:
gtnhenbr [62]3 years ago
6 0

Answer:

the answer is 320

Step-by-step explanation:

just did the instruction

zubka84 [21]3 years ago
4 0

Answer:

320

Step-by-step explanation:

To find the volume, you multiply length times width times height, which is 3x1x2, which equals six. Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.

Wow look at my full percent battery

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Step-by-step explanation:

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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9
Nastasia [14]

Answer:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

Step-by-step explanation:

Given

9 < x < 10 --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

v = x^3

For a uniform distribution, we have:

x \to U(a,b)

and

f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.

9 < x < 10 implies that:

(a,b) = (9,10)

So, we have:

f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Solve

f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.

Recall that:

v = x^3

Make x the subject

x = v^\frac{1}{3}

So, the cumulative density is:

F(x) = P(x < v^\frac{1}{3})

f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right. becomes

f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.

The CDF is:

F(x) = \int\limits^{v^\frac{1}{3}}_9 1\  dx

Integrate

F(x) = [v]\limits^{v^\frac{1}{3}}_9

Expand

F(x) = v^\frac{1}{3} - 9

The density function of the volume F(v) is:

F(v) = F'(x)

Differentiate F(x) to give:

F(x) = v^\frac{1}{3} - 9

F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}

F'(x) = \frac{1}{3}v^{-\frac{2}{3}}

F(v) = \frac{1}{3}v^{-\frac{2}{3}}

So:

f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.

8 0
3 years ago
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