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3 years ago

Describe this translation.

Mathematics

1 answer:

Aliun [14]3 years ago

3 0

Answer:

The answer is B (3 units right)

Step-by-step explanation:

You pick a point in the triangle which does not have an apostrophe. Then you see the correspondent point to it and see how much it has moved and in this case it is 3 to right therefore the answer is B.

Hint: If the apostrophe is after the letter, it shows translation therefore it is the moved shape

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In 1990, the cost of tuition at a large Midwestern university was $99 per credit hour. In 2003, tuition had risen to $268 per cr

vodka [1.7K]

Answer:

63%

Step-by-step explanation:

From the question, we aim to find the percent increase in the tuition

Given data

initial cost= $99 per credit hour

Final cost= $268 per credit hour

% increase= (Final - initial )/initial *100

substitute

% increase= (268- 99 )/268 *100

% increase= 169 /268 *100

% increase= 0.630*100

% increase= 63%

Hence the increase in the tuition from 1990 to 2003 is 63%

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:

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