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Alex
3 years ago
9

Which expession is equivalent to -3(6c + 2) + 5c

Mathematics
2 answers:
sp2606 [1]3 years ago
6 0

Answer:

-13c-6

Step-by-step explanation:

-3(6c + 2) + 5c

Distribute

-18c -6 +5c

Combine like terms

-13c-6

rjkz [21]3 years ago
6 0

Answer:

-13c -6

Step-by-step explanation:

-3 (6c) -3x2+5c

multiply 6 by -3

-18c-3x2+5c

multiply -3 by 2

-18c -6+ 5c

add 18c and 5c

-13c -6

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The Blair family was one of the first to come to the original 13 colonies (now part of the USA). They had 4 children. Assuming t
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Answer:

31.25% probability that the Blair family had at least 3 girls

Step-by-step explanation:

For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. 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.

They had 4 children.

This means that n = 4

The probability of a child being a girl is 0.5

This means that p = 0.5

Probability of at least 3 children:

P(X \geq 3) = P(X = 3) + P(X = 4)

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

P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25

P(X = 4) = C_{4,4}.(0.5)^{4}.(0.5)^{0} = 0.0625

P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125

31.25% probability that the Blair family had at least 3 girls

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3 years ago
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Step-by-step explanation:

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