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AlexFokin [52]
3 years ago
11

a scalene right triangle has a hypotenuse with a length of 50 cm. The length of one leg is twice the length of the other. What a

re the lengths of the legs of the triangle to the nearest tenth of a centimeter. Show your work.
Mathematics
1 answer:
zheka24 [161]3 years ago
4 0
Let
x------------> <span>The length of one leg
</span>y------------> <span>The length of second leg
</span>
we know that
H=50 cm
x=2y
<span>applying the Pythagorean theorem

H</span>²=x²+y²--------> 50²=(2y)²+y²---------> 2500=4y²+y²-----> y²=2500/5
y=√500--------> y=22.4 cm
<span>so
</span>x=2y--------> x=2*22.4--------> x=44.8 cm
<span>
the answer is
</span><span>the lengths of the legs of the triangle are
</span>44.8 cm
and
22.4 cm
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Seventy-two percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of th
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Answer:

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c) 0.064 = 6.4% probability that 7 of them are discovered.

Step-by-step explanation:

For itens a and b, we use conditional probability.

For item c, we use the binomial distribution along with the conditional probability.

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

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.

a) If it has an emergency locator, what is the probability that it will not be discovered?

Event A: Has an emergency locator

Event B: Not located.

Probability of having an emergency locator:

66% of 72%(Are discovered).

20% of 100 - 72 = 28%(not discovered). So

P(A) = 0.66*0.72 + 0.2*0.28 = 0.5312

Probability of having an emergency locator and not being discovered:

20% of 28%. So

P(A cap B) = 0.2*0.28 = 0.056

Probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.056}{0.5312} = 0.105

0.105 = 10.5% probability that it will not be discovered if it has an emergency locator.

b) If it does not have an emergency locator, what is the probability that it will be discovered?

Probability of not having an emergency locator:

0.5312 of having. So

P(A) = 1 - 0.5312 = 0.4688

Probability of not having an emergency locator and being discovered:

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P(A \cap B) = 0.34*0.72 = 0.2448

Probability:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2448}{0.4688} = 0.522

0.522 = 52.2% probability that it will be discovered if it does not have an emergency locator.

c) If we consider 10 light aircraft that disappeared in flight with an emergency recorder, what is the probability that 7 of them are discovered?

p is the probability of being discovered with the emergency recorder:

0.5312 probability of having the emergency recorder.

Probability of having the emergency recorder and being located:

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Probability of being discovered, given that it has the emergency recorder:

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This question asks for P(X = 7) when n = 10. So

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

P(X = 7) = C_{10,7}.(0.8946)^{7}.(0.1054)^{3} = 0.064

0.064 = 6.4% probability that 7 of them are discovered.

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