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lisov135 [29]
3 years ago
5

I will give the BRAINIEST!!! Please help me and dont just guess this is hard.

Mathematics
1 answer:
Verdich [7]3 years ago
5 0

Answer:

I do not know the answer. Ask "LucasPeed" He is good at those answers.

Step-by-step explanation:

If you need any help ask Brainly. Have nice day!

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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
How is 2.5/3=75% equivalent
horsena [70]

Answer:

its equivalent because 75% equals to 75 over one hundred = 75/100 so when you divide 75 by 3 you get 25 but then you have to divide 25 by 10 you get 2.5

Step-by-step explanation:

4 0
3 years ago
Find the 12th term given the arithmetic sequence: an = -5n +32
Andrej [43]

Step-by-step explanation:

step 1. a12 = -5(12) + 32

step 2. a12 = -60 + 32

step 3. a12 = -28.

7 0
3 years ago
A wooden plank of length x is cut into 8 sections such that each section is at most
Levart [38]
Well you take the length of each section, which is "blank" rn, and multiply into 8? I don't know what will the meaning of "at most" be effective here, but if it's normal calculations, x = "blank" x 8
5 0
3 years ago
The length of each side of a cube is x + 1 inches. What is the volume of the cube in cubic inches?
castortr0y [4]

Answer:

V = x³ + 3x² + 3x + 1 in³

Step-by-step explanation:

The volume (V) of a cube is calculated as

V = s³ ( where s is the side ) , then

V = (x + 1)³

   = (x + 1)(x + 1)(x + 1) ← expand second pair using FOIL

   = (x + 1)(x² + 2x + 1) ← distribute

   = x³ + 2x² + x + x² + 2x + 1 ← collect like terms

   = x³ + 3x² + 3x + 1

3 0
3 years ago
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