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Nady [450]
3 years ago
7

Describe the pattern. Then find the next two terms of the sequence. 99, 88, 77, 66

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer: The pattern of the sequence is to subtract 11 from the last number.

Step-by-step explanation: The next two terms of the sequence are 55 and 44.

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A certain shop repairs both audio and video components. Let A denote the event that the next
alexdok [17]

Answer:

0.12 is the required probability.

Step-by-step explanation:

We are given the following in the question:

A: next  component brought in for repair is an audio component

B: event that the next component  is a compact disc player

B \in A

P(A)= 0.6\\P(B) =0.05

We have to evaluate:

P(B|A) = \dfrac{P(B\cap A)}{P(A)}\\P(B|A) = \dfrac{P(B)}{P(A)}\\\\P(B|A) = \dfrac{0.05}{0.6} = 0.12

0.12 is the required probability.

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3 years ago
Alyssa has fifty-quarters and dimes
Bezzdna [24]
A. 32

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4 0
2 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
I need help please and thank you
lys-0071 [83]

Answer: it’s 135

Step-by-step explanation:

4 0
3 years ago
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