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3 years ago

The volume of a sphere whose diameter is 18 cm is

Mathematics

1 answer:

prohojiy [21]3 years ago

5 0

Answer:

= 3052.08 cm^3

Step-by-step explanation:

Volume of a sphere is given by

V = 4/3 pi r^3

We know the diameter is 18

d = 2r

18 = 2r

Divide each side by 2

18/2 =2r/2

9 =r

V = 4/3 pi 9^3

V = 4/3 pi (729)

V = 972 pi

Approximating pi by 3.14

V =972 (3.14)

= 3052.08 cm^3

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Find (f+g)(x) if f(x)=3x+10x and g(x)=5x-3

otez555 [7]

3x-5x+3 Answer:

Step-by-step explanation:

7 0

4 years ago

Please help!!!!!!!!

Lynna [10]

Answer: the number, converted to base 10 is 884

Step-by-step explanation:

The table gives symbols for a base 12 numerical system and it is called Caidoz numbers.

Looking at the table, the corresponding Caidozian number that we considering is 618 to base 12. Converting 618 to base 10, it becomes

8 × 12^0 = 8 × 1 = 8

1 × 12^1 = 1 × 12 = 12

6 × 12^2 = 6 × 144 = 864

Therefore, 618 to base 12 to 618 to base 10 would be

8 + 12 + 864 = 884

8 0

3 years ago

The product of 4.7 and 6.5 equals 30.55 what is the product of 4.7 and 0.65 4.7 and 65

lora16 [44]

Your answer is, 198.575 math might be off so let me check.

8 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

4 years ago

1a.) Given the equation 5+ 6n-4n-15, explain how to get the answer n10.

Savatey [412]

Answer:

Step-by-step explanation:

combine like terms (6n-4n)=(5-15)

2n=-20

n=20/2

n=10

7 0

3 years ago