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DedPeter [7]
3 years ago
11

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired

, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly three will end up being replaced under warranty? (Round your answer to three decimal places.)
Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer: 0.034

Step-by-step explanation:

Given : P(Submitted under warranty)= 0.20

P(Replaced  | Submitted under warranty)=0.40

P(Replaced and Submitted under warranty )= P(Submitted under warranty)×P(Replaced  | Submitted under warranty)

=0.20\times0.40=0.08

Let x be the number of telephones will end up being replaced under warranty.

Total telephones purchased : n= 10

Using binomial probability formula : P(X)=^nC_xp^x(1-p)^{n-x}

i.e. The probability that exactly three will end up being replaced under warranty will be :-

P(X=3)=^{10}C_3(0.08)^3(1-0.08)^{10-3}\\\\=\dfrac{10!}{3!(10-3)!}(0.08)^3(0.92)^7\\\\=0.03427409518\approx0.034 [Rounded to three decimal places. ]

Hence, the probability that exactly three will end up being replaced under warranty : 0.034

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Answer:

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Step-by-step explanation:

Area of sector=pi*r^2*(theta/360) = pi*r^2*(45/360)

18*pi=pi*r^2*(1/8), r=12. Circumference is 24*pi

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3 years ago
70=(4x+36)/2 figure this out please
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Answer:

\boxed{\bold{\huge{\boxed{x = 26}}}}

Step-by-step explanation:

70 = \frac{4x+36}{2}

<u>Multiplying both sides by 2</u>

70 * 2 = 4x+36

140 = 4x+36

<u>Subtracting 36 to both sides</u>

140 - 36 = 4x

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OR

x = 26

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3 years ago
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3/10 is represented as 30%.
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The J.O. Supplies Company buys calculators from a non-US supplier. The probability of a defective calculator is 10 percent. If 3
RSB [31]

Answer:

There is a 24.3% probability that one of the calculators will be defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability of a defective calculator is 10 percent.

This means that p = 0.1

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243

There is a 24.3% probability that one of the calculators will be defective.

3 0
3 years ago
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