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My name is Ann [436]
3 years ago
10

4. A sequence of transformations was applied to an equilateral triangle in a coordinate plane.

Mathematics
1 answer:
forsale [732]3 years ago
8 0

Step-by-step explanation:

A It must be an equilateral triangle with the same side lengths as the original triangle

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(3c² + 2d)(–5c² + d)-------->(3c²)*(-5c²)+(3c²)*(d)+(2d)*(-5c²)+(2d)*(d)
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The probability that a person in the United States has type B​+ blood is 12​%. Three unrelated people in the United States are s
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Answer:

The probability that all three have type B​+ blood is 0.001728

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have type B+ blood, or they do not. The probability of a person having type B+ blood is independent of any other person. 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.

The probability that a person in the United States has type B​+ blood is 12​%.

This means that p = 0.12

Three unrelated people in the United States are selected at random.

This means that n = 3

Find the probability that all three have type B​+ blood.

This is P(X = 3).

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

P(X = 3) = C_{3,3}.(0.12)^{3}.(0.88)^{0} = 0.001728

The probability that all three have type B​+ blood is 0.001728

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