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4 years ago

Can you expand 2(3c-2)

Mathematics

2 answers:

slamgirl [31]4 years ago

7 0

I think it would be 2c-4

Nataly [62]4 years ago

3 0

Like when we expend it we should multiply 2 by 3c and then multiply by 2
2(3c-2) = 6c-4 (it is the answer)

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Step-by-step explanation:

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3 years ago

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2 years ago

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7.109) The leading brand of dishwasher detergent has a 30% market share. A sample of 25 dishwasher detergent customers was taken

Rina8888 [55]

Answer:

0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they choose the leading brand, or they do not. The probability of a customer choosing the leading brand is independent of any other customer, which means that 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.

The leading brand of dishwasher detergent has a 30% market share.

This means that $p = 0.3$

A sample of 25 dishwasher detergent customers was taken.

This means that $n = 25$

a. What is the probability that 10 or fewer customers choose the leading brand?

This is:

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 0) = C_{25,0}.(0.3)^{0}.(0.7)^{25} = 0.0001$

$P(X = 1) = C_{25,1}.(0.3)^{1}.(0.7)^{24} = 0.0014$

$P(X = 2) = C_{25,2}.(0.3)^{2}.(0.7)^{23} = 0.0074$

$P(X = 3) = C_{25,3}.(0.3)^{3}.(0.7)^{22} = 0.0243$

$P(X = 4) = C_{25,4}.(0.3)^{4}.(0.7)^{21} = 0.0572$

$P(X = 5) = C_{25,5}.(0.3)^{5}.(0.7)^{20} = 0.1030$

$P(X = 6) = C_{25,6}.(0.3)^{6}.(0.7)^{19} = 0.1472$

$P(X = 7) = C_{25,7}.(0.3)^{7}.(0.7)^{18} = 0.1712$

$P(X = 8) = C_{25,8}.(0.3)^{8}.(0.7)^{17} = 0.1651$

$P(X = 9) = C_{25,9}.(0.3)^{9}.(0.7)^{16} = 0.1336$

$P(X = 10) = C_{25,10}.(0.3)^{10}.(0.7)^{15} = 0.0916$

Then

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0001 + 0.0014 + 0.0074 + 0.0243 + 0.0572 + 0.1030 + 0.1472 + 0.1712 + 0.1651 + 0.1336 + 0.0916 = 0.9021$