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2 years ago

The totals of the rows and columns of a two way table are called marginal distributions.

2 answers:

5 0

The answer is option B "false". A marginal distribution is a set of random variables. (not on a column or graph) Also a marginal distribution is a measurement used in the fields of statistics and probability.

Hope this helps!

Ber [7]2 years ago

3 0

Answer:

True!

Step-by-step explanation:

just did it on apex

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During the day,Sam spent $4.85 on lunch.he also bought 2 books for $7.95 each. At the end of the day we had $8.20 left how much

Vlada [557]

20.75 because if you add 4.85+7.95+7.95 add that you get 20.75

7 0

2 years ago

Simplify 5^2•(4+2^3)

Marta_Voda [28]

Remember to follow PEMDAS, and left -> right rule.

First, solve the Parenthesis:

(4 + 2^3)

Solve the exponent.

2^3 = 2 x 2 x 2 = 4 x 2 = 8

4 + 8 = 12

_______________________________________________________________

(5^2) x (12)

5^2 = 5 x 5 = 25

25 x 12

_______________________________________________________________

Multiply

25 x 12 = 300

________________________________________________________________

300 is your answer

________________________________________________________________

~Rise Above the Ordinary

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2 years ago

I put 80 POINTS into question please answer

anzhelika [568]

The answer is -0.4. hope this helps

8 0

2 years ago

Read 2 more answers

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kvasek [131]

Hello,

Answer D

x²/25+y²/16=1

5 0

2 years ago

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

The question is incomplete:

"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

2 years ago