All categories

2 years ago

A solid plastic ball is a sphere with a radius of 8 inches. How much plastic does it take to make one ball

Mathematics

1 answer:

mihalych1998 [28]2 years ago

7 0

Answer:

2144.66in³

Step-by-step explanation:

Radius = 8 inches.

To determine the amount of sphere to make the ball, we'll have to find its volume.

Volume of a sphere = 4/3 * Πr³

V = 4/3*Πr³

Π = 22/7

V = 4/3 * 22/7 * 8³

V = 2144.66inches³

V = 2144.66in³

The amount of plastic required is 2144.66in³.

You might be interested in

If a sphere has a circumference of 138.16 in., what is the surface area of the sphere?

BigorU [14]

The surface area is 6075.96 in^3

5 0

3 years ago

.What is the value of the expression below when x=9? Enter the answer as NUMBER only? 1/3 x^2+4

belka [17]

Step-by-step explanation:

1 *9^2 + 4

1 * 81 + 4

1 * 85

85

=28

5 0

2 years ago

Which shows the numbers One-tenth, 0.9, Three-tenths in order from least to greatest?

liq [111]

A
If you put them all into decimal form you get .1, .3, .9

6 0

2 years ago

Need help please answer asap

Vikki [24]

Answer:

$2x = x + 15$

Step-by-step explanation:

Angle AGH and GHD are corresponding angles, meaning they are equal. So in order to find the equation that can be used to solve for x we need to equate both angles AGH and GHD to each other.

$m < AGH\: \: \: = \: \: \: m < GHD \\ (x + 15) = (2x) \\ 15 = 2x - x \\ 15 = 1x \\ 15 = x$

They did not ask us to solve for x but I did it in case, however we can see that the answer we got corresponds with

$(1) \:2x = x + 15$

4 0

1 year ago

Read 2 more answers

A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain area who favor m

Dominik [7]

Answer:

This means that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The width of the interval is:

$W = 2z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this sample:

Two 90% intervals, with different lenghts. So both have the same values for z an $\pi$

Interval A:

9000 viewers.

So the width is

$W_{A} = 2z\sqrt{\frac{\pi(1-\pi)}{9000}}$

Interval B:

100 viewers

So the width is

$W_{B} = 2z\sqrt{\frac{\pi(1-\pi)}{1000}}$

Relationship between the widths:

$R = \frac{W_{A}}{W_{B}} = \frac{2z\sqrt{\frac{\pi(1-\pi)}{9000}}}{2z\sqrt{\frac{\pi(1-\pi)}{1000}}} = \frac{\sqrt{1000}}{\sqrt{9000}} = \frac{\sqrt{1}}{\sqrt{9}} = \frac{1}{3}$

This means that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

8 0

3 years ago