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2 years ago

HELP PLS I NEED help this is to get my grade up

Mathematics

2 answers:

pickupchik [31]2 years ago

8 0

Answer: Around 10 days. The real answer is 10 2/3.

CaHeK987 [17]2 years ago

6 0

Answer:

10 and 2-third days.

Convert mixed number to improper fraction

Initial equation

$\frac{176}{9} / \frac{11}{6}$

Next, multiply the numerators by the numerators, and do the same for the denominators. Flip the second number to 6/11

$\frac{176 * 6}{9 * 11}$

$= \frac{1056}{99}$

Simplify.

$\frac{1056}{99} -- > 10\frac{2}{3}$

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4 years ago

A rectangle has perimeter 20 m . Express the area of the rectangle as a function of the length of ones its side.

kondor19780726 [428]

Answer:

A = 10W - W²

Step-by-step explanation:

Perimeter of a rectangle = 2(L+W)

Area of a rectangle =LW

L is the length

W is the width

Given

Perimeter = 20m

20 = 2 (L+W)

10 = L+W

L = 10 - W ....2

Substitute equation 2 into the formula for calculating the area

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3 years ago

I Need help with Number 2 And 4 asap plz

oksian1 [2.3K]

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4 years ago

The combined land area of the countries A and B is 131,973 square kilometers. Country A is larger by 473 square kilometers. Dete

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x+y=131973

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4 years ago

Let X denote the length of human pregnancies from conception to birth, where X has a normal distribution with mean of 264 days a

Kaylis [27]

Answer:

Step-by-step explanation:

Hello!

X: length of human pregnancies from conception to birth.

X~N(μ;σ²)

μ= 264 day

σ= 16 day

If the variable of interest has a normal distribution, it's the sample mean, that it is also a variable on its own, has a normal distribution with parameters:

X[bar] ~N(μ;σ²/n)

When calculating a probability of a value of "X" happening it corresponds to use the standard normal: Z= (X[bar]-μ)/σ

When calculating the probability of the sample mean taking a given value, the variance is divided by the sample size. The standard normal distribution to use is Z= (X[bar]-μ)/(σ/√n)

a. You need to calculate the probability that the sample mean will be less than 260 for a random sample of 15 women.

P(X[bar]<260)= P(Z<(260-264)/(16/√15))= P(Z<-0.97)= 0.16602

b. P(X[bar]>b)= 0.05

You need to find the value of X[bar] that has above it 5% of the distribution and 95% below.

P(X[bar]≤b)= 0.95

P(Z≤(b-μ)/(σ/√n))= 0.95

The value of Z that accumulates 0.95 of probability is Z= 1.648

Now we reverse the standardization to reach the value of pregnancy length:

1.648= (b-264)/(16/√15)

1.648*(16/√15)= b-264

b= [1.648*(16/√15)]+264

b= 270.81 days

c. Now the sample taken is of 7 women and you need to calculate the probability of the sample mean of the length of pregnancy lies between 1800 and 1900 days.

Symbolically:

P(1800≤X[bar]≤1900) = P(X[bar]≤1900) - P(X[bar]≤1800)

P(Z≤(1900-264)/(16/√7)) - P(Z≤(1800-264)/(16/√7))

P(Z≤270.53) - P(Z≤253.99)= 1 - 1 = 0

d. P(X[bar]>270)= 0.1151

P(Z>(270-264)/(16/√n))= 0.1151

P(Z≤(270-264)/(16/√n))= 1 - 0.1151

P(Z≤6/(16/√n))= 0.8849

With the information of the cumulated probability you can reach the value of Z and clear the sample size needed:

P(Z≤1.200)= 0.8849

$Z= \frac{X[bar]-Mu}{Sigma/\sqrt{n} }$