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Zigmanuir [339]
3 years ago
12

The hypotenuse of right triangle ABC, line segment AC, measures 13 cm. The length of line segment BC is 5 cm.

Mathematics
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

Step-by-step explanation:

Pythagorean theorem: In a right angle triangle, the square of hypotenuse is equal to the sum of squares of other two sides

BC² + AB² = AC²

5² + AB² = 13²

25 + AB² = 169

        AB² = 169 - 25

        AB² = 144

       AB = √144 = √12*12

AB = 12cm

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Students are painting a mural. So far the mural is painted 1/4 blue 2/8 red and 3/12 green. Use fraction strips to determine how
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If you make the denominators all the same, the fractions are 3/12 , 3/12 , and 3/12. Then by simply adding them together you get that the mural is 9/12 painted. that then simplifies to  3/4 or 75%

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3 years ago
What is 13(y+7)=3(y−1)
artcher [175]

Answer: Y= -47/5

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Based on your understanding of sound waves,
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3 years ago
Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete
icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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)!}

In this question:

Fleet of 17 means that N = 17

4 are carrying nucleas weapons, which means that k = 4

9 are destroyed, which means that n = 9

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

P(X > 1) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294

P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412

P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0
3 years ago
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