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kow [346]
3 years ago
9

Jay purchased tickets for a concert over the internet. To place an order, a handling charge of 5$ per ticket is charged. GST of

5% was also charged on ticket price and handling charges. If the total charge for 2 tickets is 201.16, what is the price per ticket.
Answer is 90.79 but I need an explanation on how they got that answer thanks
Mathematics
1 answer:
mr Goodwill [35]3 years ago
5 0
Well work it out, then you should get 90.71

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Find the value of this expression if x =7<br> x² +6/x - 2
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Hello I am new to Brainly Hope it will help me​
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Nina can type 128 words is 4
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3 years ago
Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, the chance a homeowner is ins
ollegr [7]

Answer:

P(X = 0) = 0.1897

P(X = 1) = 0.3910

P(X = 2) = 0.3021

P(X = 3) = 0.1038

P(X = 4) = 0.0134

Step-by-step explanation:

For each owner, there are only two possible outcomes. Either they are insured against an earthquake, or they are not. The probability of a homeowner being insured against an earthquake is independent of other homeowners. So the binomial probability distribution is used 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.

Suppose that in one metropolitan area, the chance a homeowner is insured against an earthquake is 0.34.

This means that p = 0.34

A sample of four homeowners are to be selected at random.

This means that n = 4

(a) Find the probability mass function of X. (Round your answers to four decimal places.)

The pmf is the probability of each outcome.

So

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

P(X = 0) = C_{4,0}.(0.34)^{0}.(0.66)^{4} = 0.1897

P(X = 1) = C_{4,1}.(0.34)^{1}.(0.66)^{3} = 0.3910

P(X = 2) = C_{4,2}.(0.34)^{2}.(0.66)^{2} = 0.3021

P(X = 3) = C_{4,3}.(0.34)^{3}.(0.66)^{1} = 0.1038

P(X = 4) = C_{4,4}.(0.34)^{4}.(0.66)^{0} = 0.0134

8 0
3 years ago
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