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Zina [86]
3 years ago
15

I am paying $60.00 a month for my service. I would like to upgrade to the $80.00 service package because my new employer offers

a 20% discount with your company. What would be the cost difference compared to what I am paying now if I upgraded?"
Mathematics
2 answers:
Irina18 [472]3 years ago
8 0

Answer:

that would be 16 dollars

Step-by-step explanation:

80.00 multiplied by .2 equals 16

Fed [463]3 years ago
4 0

Answer:

The cost difference is $4.

Step-by-step explanation:

Consider the provided information.

I am paying $60 a month for my service.

I would like to upgrade to the $80.00 service package because my new employer offers a 20% discount with your company.

If i get 20% of discount that means i need to pay only 80% of the original amount.

80% of 80 can be calculated as:

\frac{80}{100}\times 80

0.8\times 80=64

Therefore, now i need to pay $4 extra for the upgrade service.

Hence, the cost difference is $4.

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6. Assume that a component passes a test is 0.85 and that components perform independently. What is the probability that the thi
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Answer:

3.90% probability that the third failure will occur on the tenth component tested

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it fails, or it does not fail. Components perform independently. So we use the binomial probability distribution to solve this question.

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.

Assume that a component passes a test is 0.85

So they fail with probability of p = 1 - 0.85 = 0.15

What is the probability that the third failure will occur on the tenth component tested

First 9 components: Two failures, that is, P(X = 2) when n = 9.

10th component: Failure with probability 0.15.

So

P = 0.15P(X = 2)

In which

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

P(X = 2) = C_{9,2}.(0.15)^{2}.(0.85)^{7} = 0.2597

So

P = 0.15P(X = 2) = 0.15*0.2597 = 0.0390

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