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ruslelena [56]
2 years ago
5

Sean's Cafe has regular coffee and decaffeinated coffee. This morning, the cafe served 18 coffees in all, 9 of which were regula

r. What percentage of the coffees were regular?
Mathematics
2 answers:
erastovalidia [21]2 years ago
8 0

Answer:

50% of the regular coffee's were regular

stiv31 [10]2 years ago
7 0
50%
Explanation:
multiply 9 • 100
the divide 900 into 18
it should give you 50
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3 years ago
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Shayla is tracking the weight of the smallest puppy in a litter. Use the table to write a linear function that models the puppy'
Pani-rosa [81]

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y=3x+3

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4 years ago
What is the value of s in the equation s-2=18
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7 0
3 years ago
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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

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3 years ago
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