(4 ten thousands 8hundreds) divided by 10

sashaice [31]

The answer I got is 8

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%282%20-%206x%29%20-%204%28x%20%20%2B%20%20%5Cfrac%7B3%7D%7B2%7D%29%20

Art [367]

Answer:

x=-2

Step-by-step explanation:

1/2(2-6x)-4(x+3/2)=-(x-3)+4

(1-3x)-4x-6=-x+3+4

-7x-5=-x+7

Tranpose

-7x+x=5+7

-6x=12

Divide by -6

x=-2

4 0

3 years ago

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:

brainly.com/question/24188394

#SPJ4

3 0

2 years ago

The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye

trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .