SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!

What are two ways to find the slope of a line?

DedPeter [7]

Answer:

m = (y - y)/(x - x)

Rise/run

Step-by-step explanation:

4 0

3 years ago

Listed below are the top 10 annual salaries (in millions of dollars) of TV personalities. Find the range, variance, and stand

Fed [463]

Answer:

1. $29.4 million

2. 145.24

3. 145.24

4. 12.05

5. D. No, because the sample is not representative of the whole population.

Step-by-step explanation:

1. The computation of range of the sample data is shown below:-

Range of the sample data = Maximum value - Minimum value

= 42 - 12.6

= $29.4 million

2. For the computation of variance of the sample data first, we need to find out the mean which is shown below:-

Mean = $\frac{(42 + 40 + 39 + 32 + 21 + 18 + 16 + 14 + 13.7 + 12.6) }{10}$

= 24.83

The Variance of the sample data

= $\frac{(42 - 24.83)^2 + ..... + (12.6 - 24.93)^2}{10 - 1}$

= $\frac{1307.161}{9}$

= 145.24

3. The computation of the standard deviation of the sample data is shown below:-

The Standard deviation of the sample

= $\sqrt{145.24}$

= $12.05155592

or

$12.05

4. No, as the sample does not represent the whole population.

8 0

3 years ago

(please help soon) Point Q is plotted on the coordinate grid. Point P is at (20, −30). Point R is vertically above point Q. It i

Triss [41]

Step-by-step explanation:

So, the distance between the points P and Q is 4 units.

8 0

3 years ago

*brilliant question*<br> Given that a= 2 to the power of M, express the following in terms of A

worty [1.4K]

Answer:

A = 1/2a^4

B= 2/a^2

Step-by-step explanation:

a= 2^(m)

A =1/2*16^m= 1/2*(2^m)^4 =1/2 a^4
B =2/(∛64)^m = 2/(2^6/3)^m = 2/2^2m = 2/a^2

4 0

3 years ago

In T-ball, the distance to each successive base is 50 feet. If the distance from home plate to the pitcher’s mound is 38 feet, h

JulijaS [17]

Answer:

<h2>The distance from the pitcher's mound and to second base is 37.99 approximately.</h2>

Step-by-step explanation:

The diamond is a square, which in this case has 50 feet long each side, and from home to pitcher is 38 feet. Notice that home is a vertex of the square and the pitcher's mound is the intersection of the diagonals, where they cut half.

We can find the distance from the pitcher to first base using Pythagorean's Theorem, where 50 feet is the hypothenuse.

$50^{2} =38^{2}+x^{2}\\x^{2}=50^{2}-38^{2}\\x=\sqrt{2500-1444}\\ x=\sqrt{1056}\\ x \approx 32.5 \ ft$

Therefore, the distance from the pitcher to first base is 32.5 feet, approximately.

Now, we can use again Pythagorean's Theorem to find the distance from pitcher to second base, where the hypothenuse is 50 feet.

$50^{2}=32.5^{2}+y^{2}\\y^{2}=50^{2}-32.5^{2}\\y=\sqrt{2500-1056.25}\\ y =\sqrt{1443} \approx 37.99$

Therefore, the distance from the pitcher's mound and to second base is 37.99 approximately.

<em>(this results make sense, because the diagonals of a square intersect at half, that means all bases have the same distance from pitcher's mound, so the second way to find the distance asked in the question is just using theory)</em>

8 0

4 years ago