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Vadim26 [7]
2 years ago
14

The side lengths of a triangle are 5,12 and 13. is this right triangle

Mathematics
2 answers:
Ahat [919]2 years ago
3 0

Answer:

<h2>Yes, because 5²+12²=13²</h2>

Step-by-step explanation:

Yes

Citrus2011 [14]2 years ago
3 0

Answer: B.) Yes, because 5²+12²=13²

Step-by-step explanation: hope that helps!(:

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we can rewrite this as a fraction. we will use 1/10s because 80 can be divided by 8 which is in the problem. we then have to find out what two numbers add up to 8 but one is 3 times the other. 6 and 2 fit into this problem so we plug it into the equation and we write it as this 6x+4x=80 where x is 8. we get 48 and 32.

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Read 2 more answers
Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems o
NISA [10]

Answer:

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

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}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so \pi = 0.8.

(a) What is the probability she gets an A?

There are four problems, so n = 4

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find P(X \geq 3)

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

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

P(X = 3) = C_{4,3}.(0.80)^{3}.(0.2)^{1} = 0.4096

P(X = 4) = C_{4,4}.(0.80)^{4}.(0.2)^{0} = 0.4096

P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so n = 3

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find P(X \geq 2)

P(X \geq 3) = P(X = 2) + P(X = 3)

P(X = 2) = C_{3,2}.(0.80)^{2}.(0.2)^{1} = 0.384

P(X = 4) = C_{3,3}.(0.80)^{3}.(0.2)^{0} = 0.512

P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896

If she gets the first problem correct, there is an 89.6% probability that she gets an A.

3 0
3 years ago
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