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Anastaziya [24]
2 years ago
8

SOMEONE PLEASE HELP ME PLEASE SOMEONE PLEASE HELP ME WITH MY HOMEWORK ILL DO ANYTHING FOR HELP

Mathematics
1 answer:
Sergeeva-Olga [200]2 years ago
4 0

Answer:

The answer is 8

Step-by-step explanation:

If this answer helped you then please consider marking this as brainliest and like this respone :)

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The answer to this picture
stepan [7]

Answer:

divide it bro

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Mary made 1/5 of a batch of cookies in 1/10th of an hour, how many batches of cookies can she make in one hour? This is unit rat
barxatty [35]

Answer:

1/5 ÷ 1/10 = 10/5

Step-by-step explanation:

7 0
3 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

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2 years ago
How do I solve this equation using long polynomial division?
gulaghasi [49]
Oh okie oh gosh lol lol I don’t know
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3 years ago
Suppose SAT scores among college students are normally distributed with a mean of 500 and a standard deviation of 100. If a stud
bagirrra123 [75]

Answer:

700-500/500=2

7 0
3 years ago
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