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2 years ago

Help I will mark you brainly

Mathematics

2 answers:

algol [13]2 years ago

7 0

Answer:

x = 20

y = 10√3 or 17.3

Step-by-step explanation:

This is a special 30° 60° 90° right triangle

In this special triangle if the side length that sees 30° is represented by x and its given as 10 the side length that sees 90° would be twice of it so it's represented by 2x = 20 and the side length that sees 60° is x√3 = 10√3

Ludmilka [50]2 years ago

5 0

Answer:

Hi there!!

The answer would be x=20 and y= 10 root 3 or on decimal it's 17.32.

explanation look in picture, alright. 

I hope it will help u....

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The probability that Barry Bonds hits a home run on any given at-bat is 0.16, and each at-bat is independent.

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a) 0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.

b) The expected number of at-bats until the next home run is 6.25.

Step-by-step explanation:

For each at bat, there are two possible outcomes. Either it is a home run, or it is not. The probability of an at bat resulting in a home run is independent of any other at-bat, which means that the binomial probability distribution is used to solve this question.

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

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

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The probability that Barry Bonds hits a home run on any given at-bat is 0.16

This means that $p = 0.16$

Part A: What is the probability that the next home run will be on his fifth at-bat?

0 on his next 4(P(X = 0) when n = 4)

Home run on his 5th at-bat, with 0.16 probability. So

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

$P(X = 0) = C_{4,0}.(0.16)^{0}.(0.84)^{4} = 0.49787136
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0.49787136 *0.16 = 0.0797.

0.0797 = 7.97% probability that the next home run will be on his fifth at-bat.

Part B: What is the expected number of at-bats until the next home run?

The expected number of trials for n successes is given by:

$E = \frac{n}{p}$

In this question, $n = 1, p = 0.16$ . So

$E = \frac{1}{0.16} = 6.25$

The expected number of at-bats until the next home run is 6.25.

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Help I will mark you brainly​

Help I will mark you brainly