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1 year ago

Surface area of hemisphere with diameter of 18

Mathematics

1 answer:

saul85 [17]1 year ago

6 0

Answer:

In terms of pi: 243 $\pi$

in decimal form: 763.407

Step-by-step explanation:

to find surface area of a hemisphere the formula is,

$3\pi r^{2}$

diameter (d) = 18

radius = d / 2 = 18 / 2

radius = 9

$3\pi (9)^{2} \\3\pi (81)\\243\pi \\OR\\763.407$

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Can someone help me please!!

Kisachek [45]

9514 1404 393

Answer:

AE = CE = 23; BE = DE = 20

Step-by-step explanation:

Put the values of the variables in their place and do the arithmetic.

AE = 2u+5 = 2(9) +5 = 23

BE = 6v-1 = 6(3.5) -1 = 20

CE = 3u-4 = 3(9) -4 = 23

DE = 8v-8 = 8(3.5) -8 = 20

The diagonals cross at their midpoints, so the quadrilateral is a parallelogram.

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2 years ago

The school office ordered 24 boxes of pens. Their total cost was $19152. What was the cost of 1 box of pens?

aleksley [76]

Answer:

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3 years ago

Find the cube root of 19683 by prime factorization method. Show steps.

ANTONII [103]

Answer:

Step-by-step explanation:

Given number:

19683

Prime factors of 19683:

19683 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 = 3^9

Cube root:

∛19683 = ∛3^9 = 3^3 = 27

3 0

3 years ago

Read 2 more answers

The amount of money Kelly earns is proportional to the number of hours she works. Kelly earns $32.40 working 4 hours. Create an

UkoKoshka [18]

She earns 32.40 for working 4 hrs.....32.40/4 = 8.10 per hr

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3 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

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3 years ago