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uranmaximum [27]

3 years ago

10

Cube A is inscribed in sphere B, which is inscribed in cube C. If the sides of cube A have length 4, what is the volume of cube

C?

1 answer:

Aleks04 [339]3 years ago

4 0

Answer:

The volume of cube C is

$V=192\sqrt{3}\ units^3$

Step-by-step explanation:

step 1

Find the diameter of sphere B

we know that

When a cube is inscribed in a sphere, the long diagonal of the cube is a diameter of the sphere

Let

L ----> the length side of cube A

d ----> the diagonal of the base of cube A

D ---> the long diagonal of cube A

Find the diagonal of the base of cube A

Applying the Pythagorean Theorem

$d^2=L^2+L^2$

we have

$L=4\ units$

substitute

$d^2=4^2+4^2\\d^2=32\\d=4\sqrt{2}\ units$

Find the long diagonal of cube A

Applying the Pythagorean Theorem

$D^2=d^2+L^2$

substitute

$D^2=32+4^2\\D^2=48\\D=4\sqrt{3}\ units$

step 2

we know that

If sphere B is inscribed in cube C, then the length side of cube C is equal to the diameter of sphere B

Let

c ----> the length side of cube C

we have that

$c=4\sqrt{3}\ units$

The volume of cube C is equal to

$V=c^3$

substitute

$V=(4\sqrt{3})^3$

$V=192\sqrt{3}\ units^3$

You might be interested in

Find the measure of each side indicated. Round to the nearest tenth. Picture included below.

skad [1K]

Step-by-step explanation:

Use the sine and cosine trig functions

21) cos(52)=13/x

x=13/cos(52)

x=21.1

22) sin(75)=6/x

x=6/sin(75)

x=6.2

4 0

3 years ago

Assume that there is a 4% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

kap26 [50]

Answer:

a) 99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b) 99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

4% rate of disk drive failure in a year.

This means that 96% work correctly, $p = 0.96$

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 2$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984$

99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 4$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744$

99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

7 0

3 years ago

Plsss help me Brainlist!!

gulaghasi [49]

I believe the answer to this is:

Mean: <u>8.7</u> minutes

Median: <u>9</u> minutes

Hope this helps! :D

5 0

3 years ago

From 12,000 graphing calculators produced by a manufacturer, an inspector selects a random sample of 550 calculators and finds

maria [59]

Answer:

131

Step-by-step explanation:

Using proportion,

If Out of 550 calculators = 6 are defective,

Then out of 12000 calculators = ? are defective

$\frac{12000 calculators}{550calculators} * 6$

= 21.8 * 6

= 130.8 ≈ 131

∴ Approximately 131 calculators are defective

3 0

3 years ago

So what y’all know about this???

11111nata11111 [884]

Answer:

1.052

Step-by-step explanation:

Add 1 to the percent increase.

1 + 5.2% = 1 + 5.2/100 = 1 + 0.052 = 1.052

Answer: 1.052

4 0

3 years ago