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sertanlavr [38]
3 years ago
14

Stan's, marks, and wayne's ages are consecutive whole numbers. stan is the youngest, and wayne is the oldest. the sum of their a

ges is 108. find their ages.
Mathematics
1 answer:
White raven [17]3 years ago
4 0

Stan's, Mark's and Wayne's ages are 35 , 36 and 37 years respectively.

<em><u>Explanation</u></em>

Stan's, Mark's and Wayne's ages are <u>consecutive whole numbers</u> and Stan is the youngest and Wayne is the oldest.

So, lets assume that Stan's, Mark's and Wayne's ages are  n, n+1 and n+2

Given that, the sum of their ages is 108. So, the equation will be.....

n+(n+1)+(n+2)=108\\ \\ 3n+3=108\\ \\ 3n=108-3=105\\ \\ n=\frac{105}{3}=35

So, Stan's age is 35 years , Mark's age is (35+1)= 36 years and Wayne's age is (35+2)= 37 years.

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In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
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Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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