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2 years ago

11. Which is a set of similar fractions?

Mathematics

1 answer:

Juliette [100K]2 years ago

7 0

Answer:

D. 2/10, 3/10, 4/10

Step-by-step explanation:

D. is the correct answer because the fractions have a common denominator of 10, whereas the other answer choices do no.

D. has similar fractions

Good luck to you!

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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

Help please :))))))))

docker41 [41]

Answer: 35/24 is the answer for the first part only.

Use calculator soup on google search to answer the rest.

Step-by-step explanation:

3 0

3 years ago

Last month Mrs. Hamilton bought a 4-pound bag of potatoes for $9.12. This month she purchased a 6-pound bag of potatoes for $11.

kirza4 [7]

38 cents, could be wrong tho!

6 0

2 years ago

An insurance company divides its policyholders into low-risk and high-risk classes. 60% were in the low-risk class and 40% in th

AleksAgata [21]

Answer:

a). 0.294

b) 0.11

Step-by-step explanation:

From the given information:

the probability of the low risk = 0.60

the probability of the high risk = 0.40

let $C_o$ represent no claim

let $C_1$ represent 1 claim

let $C_2$ represent 2 claim :

For low risk;

so, $C_o$ = (0.80 * 0.60 = 0.48), $C_1$ = (0.15* 0.60=0.09), $C_2$ = (0.05 * 0.60=0.03)

For high risk:

$C_o$ = (0.50 * 0.40 = 0.2), $C_1$ = (0.30 * 0.40 = 0.12) , $C_2$ = ( 0.20 * 0.40 = 0.08)

Therefore:

a), the probability that a randomly selected policyholder is high-risk and filed no claims can be computed as:

$P(H|C_o) = \dfrac{P(H \cap C_o)}{P(C_o)}$

$P(H|C_o) = \dfrac{(0.2)}{(0.48+0.2)}$

$P(H|C_o) = \dfrac{(0.2)}{(0.68)}$

$P(H|C_o) = 0.294$

b) What is the probability that a randomly selected policyholder filed two claims?

the probability that a randomly selected policyholder be filled with two claims = 0.03 + 0.08

= 0.11

7 0

2 years ago

Do you need help with geometry?? Let me help<br> send me your questions and i will help you!

Vlad [161]

Answer:

I dont have a question but thanks so much for doing that!

Step-by-step explanation:

7 0

2 years ago

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