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2 years ago

PLEASE HELP!?!?!

Mathematics

1 answer:

nalin [4]2 years ago

3 0

It’s b hope that helped have a great night

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Suppose that E and F are two events and that N(E and F)=300 and N(E)=730. What is P(F|E)?

marin [14]

Answer:

$\displaystyle\frac{30}{73}$

Step-by-step explanation:

The formula of conditional probability is:

$\displaystyle P(F|E)=\frac{P(E \cap F)}{P(E)}$

And if we called N(U) the total number of possible outcomes, then remember by definition of probability:

$\displaystyle P(E \cap F)=\frac{N(E \cap F)}{N(U)}\\\displaystyle P(E)=\frac{N(E)}{N(U)}$

Then plugging them into the formula of conditional probability we get:

$\displaystyle P(F|E)=\frac{\frac{N(E \cap F)}{N(U)}}{\frac{N(E)}{N(U)}}$

Then we simplify and we get:

$\displaystyle P(F|E)=\frac{N(E \cap F)}{N(E)}$

We just plug the given info and we get:

$\displaystyle P(F|E)=\frac{300}{730}=\frac{30}{70}$

3 0

2 years ago

MATH if someone can please help that would be amazing thank you!

White raven [17]

223,200,000 should be correct

5 0

1 year ago

A $525 speaker now sells for $393.75. Find the percent of change.

antiseptic1488 [7]

Answer: 25%

Step-by-step explanation:

595 - 393.75 = 131.25

131.25 (amount of change)/ 525 (original amount) x 100%

= 25%

8 0

2 years ago

Calculate the speed of a car that went a distance of 125 miles in 2 hours<br> time.

DerKrebs [107]

Answer:

62.5

Step-by-step explanation:

speed = distance / time taken

= 125/2

=62.5miles per hour

5 0

2 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$ , and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0

2 years ago