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3 years ago

14

The sum of the digits of a three-digit number is 6. The hundreds digit is twice the units digit, and the tens digit equals to th

e sum of the other two. Find the number.

1 answer:

wlad13 [49]3 years ago

6 0

Answer:

231

Step-by-step explanation:

231

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For a repeated-measures study comparing two treatments with 12 scores in each treatment, what is the df value for the t statisti

marishachu [46]

The degree of freedom for t statistic is 11.

According to the given question.

For a repeated-measure study, comparing two treatments with 12 scores in each treatment .

So, we can say that sample size, n = 12.

We know that, when you have a sample and estimate the mean, we have

n – 1 degrees of freedom, where n is the sample size.

Therefore,

The degree of freedom for the given sample test will be

d.f = n -1

⇒ d.f = 12 - 1

⇒ d.f = 11

Hence, the degree of freedom for t statistic is 11.

Find out more informaion about degree of freedom for sample test here:

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2 years ago

Finding the Likelihood of an Event

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Answer:

0/6, 1/3, 1/6, 5/6, 1/2, 2/3, 1/2, 1/3

Step-by-step explanation:

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4 years ago

What is an equation of the line that is parallel to the one shown and passes through (6, −5)?

Ulleksa [173]

Answer:

Step-by-step explanation:

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3 years ago

How do I find the answer to the second question?

dimaraw [331]

The sentence does not make sense because a test can only be 100% not 200

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4 years ago

A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work pro

olchik [2.2K]

Answer:

0.756

Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

$P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92$

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of $P(A\cap B\cap C\cap D)$ .

If two events X and Y are independent, then

$P(X\cap Y)=P(X)\times P(Y)$

Assume the probability of one part working does not depend on the functionality of any of the other parts.

$P(A\cap B\cap C\cap D)=P(A)\times P(B)\times P(C)\times P(D)$

Substitute the given values.

$P(A\cap B\cap C\cap D)=0.93\times 0.93\times 0.95\times 0.92$

$P(A\cap B\cap C\cap D)=0.7559226$

$P(A\cap B\cap C\cap D)\approx 0.756$

Therefore, the probability that the machine works properly is 0.756.

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3 years ago