All categories

3 years ago

Work out the volume of this sphere?

Mathematics

1 answer:

Gekata [30.6K]3 years ago

6 0

Step-by-step explanation:

$v = \frac{4}{3} \pi {( \frac{19}{2} )}^{3} \\ v = \frac{4}{3} \pi \: 857.375 = 3591.364 \\ = 3591.4 \: {cm}^{3}$

You might be interested in

Estimate the area of Alberta in square miles. Show your reasoning

ella [17]

Answer: About $278,250\ mi^2$

Step-by-step explanation:

The missing figure is attached.

Notice in the first picture that Alberta has a complex shape.

You can calculate the area of a complex shape by decomposing it into polygons whose areas can be calculated easily.

Observe the second picture. Notice that it can be descompose into two polygons: A trapezoid and a rectangle.

The area of the trapezoid can be calcualted with the formula:

$A_t=\frac{h}{2}(B+b)$

Where "h" is the height, "B" is the long base and "b" is short base.

And the area of the rectangle can be found with the formula:

$A_r=lw$

Wkere "l" is the lenght and "w" is the width.

Then, the apprximate area of Alberta is:

$A=\frac{h}{2}(B+b)+lw$

Substituting vallues, you get:

$A=(\frac{(410\ mi-180\ mi)}{2})(760\ mi+470\ mi)+(180\ mi)(760\ im)\\\\A=141,450\ mi^2+136,800\ mi^2\\\\A=278,250\ mi^2$

Therefore, the area of of Alberta is about $278,250\ mi^2$ .

5 0

2 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 $n = 4$