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2 years ago

Will mark Brainliest! Find the volume of the composite figure

Mathematics

2 answers:

Oksanka [162]2 years ago

6 0

The volume of the composite figure is 102

nasty-shy [4]2 years ago

5 0

Answer:

Answer is 102

Step-by-step explanation:

multiply 21x16=336

18x13=234

336-234=102

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Qual e a raiz quadrada de 49

dimulka [17.4K]

Answer:

it's 7

Step-by-step explanation:

5 0

3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

$P = P_{1} + P_{2} + P_{3}$

$P_{1}$ is the probability that all three of them are 13-watt. So:

$P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277$

$P_{2}$ is the probability that all three of them are 18-watt. So:

$P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415$

$P_{3}$ is the probability that all three of them are 23-watt. So:

$P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173$

$P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865$

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

$P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245$

There is a 12.45% probability that one bulb of each type is selected.

3 0

3 years ago

A store advertises 15% off an item that regularly sells for $300.

Ilia_Sergeevich [38]

Regular price = 300

Sale = 15%

Price after sale

= 300 - 15%(300)

= 300 - 15/100 x 300

= 300 - 15 x 3

= 300 - 45

= 255

7 0

2 years ago

Find the equation of the line passing through the point (4,−1) that is parallel to the line 2x−3y=9 Find the slope of the line 2

AnnZ [28]

Answer:

Step-by-step explanation:

2x - 3y = 9

-3y = -2x + 9

$y=\frac{-2}{-3}x + \frac{9}{-3}\\\\y=\frac{2}{3}x-3\\$

Parallel lines have same slope.So,

Slope m = 2/3

(4 , -1)

Equation: y - y1 = m(x - x1)

$y-[-1]=\frac{2}{3}(x - 4)\\\\y+1=\frac{2}{3}*x - \frac{2}{3}*4\\\\y+1=\frac{2}{3}x-\frac{8}{3}\\\\y=\frac{2}{3}x-\frac{8}{3}-1\\\\y=\frac{2}{3}x-\frac{8}{3}-\frac{3}{3}\\\\y=\frac{2}{3}x-\frac{11}{3}$

b = -11/3

5 0

3 years ago

Read 2 more answers

Solve for a 1/4 (20 - 4a) = 6- a G. No Solution H. All Real Numbers I. -6 J. -6

Margaret [11]

Answer:

THE EQUATION HAS NO SOLUTION!

Step-by-step explanation:

Simplify 1/4 , (1/4 • (20 - 4a)) - (6 - a) = 0 , 3.1 Pull out like factors : 20 - 4a = -4 • (a - 5) , (5 - a) - (6 - a) = 0 , -1 = 0 , 5.1 Solve : -1 = 0, THERES NO SOLUTION MATE TRUST ME!

4 0

3 years ago