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lapo4ka [179]
3 years ago
14

What is the median for this list of numbers? 4, 14, 8, 19, 22

Mathematics
2 answers:
Setler [38]3 years ago
8 0

Answer:

the answer is 8

Step-by-step explanation:

AleksandrR [38]3 years ago
8 0

Answer:

14

Step-by-step explanation:

Big Brain

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Determine which of the following are functions. Select all that apply.
Firlakuza [10]

Answer:

I am led to believe the answers are 4 and 1

Step-by-step explanation:

I don't think its answer 2 because thou are radios and their isn't any fractions in it

I solved answer 4 it came as a fraction

Okay i looked at 1 and i did all of them they all came out as a decimal so i think that's another one

answer 3 you can't divide by 0 so i don't think so

8 0
3 years ago
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What is the value of 9 in 6543.965?​
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9 value is ====*0.9*

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5 0
2 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} }

Z*(Sigma/\sqrt{n} )= (X[bar]-Mu)

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

Sigma= \frac{(X[bar]-Mu)}{Z}*\sqrt{n}

Sigma*(\frac{Z}{(X[bar]-Mu)})= \sqrt{n}

n = (Sigma*(\frac{Z}{(X[bar]-Mu)}))^2

n = (16*(\frac{1.2}{(270-264)}))^2

n= 10.24 ≅ 11 pregnant women.

I hope it helps!

6 0
2 years ago
A laundry basket contains 8 white t-shirts and 6 black t-shirts. What is the probability of randomly picking a black t-shirt fro
MissTica

Answer:

30/182 = 15/91.

Step-by-step explanation:

<u>PROBABILITY:</u> the likelihood of an event.

There are a total of 14 shirts in the basket.

Of the 14, 8 (8/14) of them are white, and 6 (6/14) of them are black.

The probability of picking a black shirt is 6 in 14, or 6/14.

The probability of picking a white shirt is 8 in 14, or 8/14.

<u>WITHOUT REPLACING:</u> This means that you take one without putting it back. (There will be one less).

6/14 × 5/13 = 30/182.

30/182 = 15/91.

8 0
3 years ago
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