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4 years ago

-2 3/4 ÷ -1/7= help

Mathematics

1 answer:

Hitman42 [59]4 years ago

3 0

The answer is 19 $\frac{1}{4}$

First you must set up the equation

$-2\frac{3}{4}$ ÷ $-\frac{1}{7}$

First, you must make the mixed number a improper fraction

$-\frac{11}{4}$ ÷ $-\frac{1}{7}$

Then you must divide

$\frac{77}{4}$

Finally, simplify

19 $\frac{1}{4}$

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Item 4

dybincka [34]

Answer:

9440

Step-by-step explanation:

4% of 9000=360. Knowing that, we can do 9000-360 which equals 8640+800 is 9440. Hope this helps :)

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3 years ago

Solve for the angle denoted by the "?". Round to the nearest tenth.

katovenus [111]

<h3>Answer: 35.4</h3>

====================================================

Explanation:

Let x be the reference angle where the question mark is located.

Opposite this angle x is 27. The adjacent side is 38.

Use the tangent ratio to tie the two sides together

tan(angle) = opposite/adjacent

tan(x) = 27/38

x = arctan(27/38)

x = 35.3947958 .... which is approximate

x = 35.4

Be sure that your calculator is in degree mode. The notation "arctan" refers to "arctangent" which is the same as inverse tangent. Your calculator should have a button labeled $\tan^{-1}$ to help compute this.

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3 years ago

A certain test preparation course is designed to help students improve their scores on the USMLE exam. A mock exam is given at t

bogdanovich [222]

Answer:

The critical value is $T = 4.604$ .

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

Step-by-step explanation:

The first step to solve this question is finding the sample mean and sample standard deviation:

14,11,18,9,19

Sample mean is sum of all values divided by the number of values. Thus:

$\overline{x} = \frac{14+11+18+9+19}{5} = 14.2$

The sample standard deviation is the square root of the division of the sum of the subtractions squared of each value and the mean, and the number of values. Thus:

$s = \sqrt{\frac{(14-14.2)^2+(11-14.2)^2+(18-14.2)^2+(9-14.2)^2+(19-14.2)^2}{5}} = 3.868$

Confidence interval:

We have the standard deviation for the sample, and so we use the t-distribution to build the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 5 - 1 = 4

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 4.604.

The critical value is $T = 4.604$ .

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 4.604\frac{3.868}{\sqrt{5}} = 7.964$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 14.2 - 7.964 = 6.236.

The upper end of the interval is the sample mean added to M. So it is 14.2 + 7.964 = 22.164.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

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-2 3/4 ÷ -1/7= help​

-2 3/4 ÷ -1/7= help