Interpreting (1,15) with (4,60), found that (4,60) is 4 times of the coordinate (1,15).
<h3>What are coordinates?</h3>
A pair of numbers that describe the position of a point on a coordinate plane by using the horizontal and vertical distances from the two reference axes.
For interpreting we have to find the relation between (1,15) and (4,60)
We have, (1,15)
For x- coordinate 1, y- coordinate will be 15
Now,
let us take x- coordinate=2 then y- coordinate =30
again, if x-coordinate = 3 then y-coordinate = 45
lastly, if the x-coordinate =4 then y-coordinate = 60
That means (4,60) is the 4 times of the coordinate (1,15) i.e., the x-coordinate is 4 times as well as the y- coordinate is 4 times.
Learn more about coordinates here:
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This problem is accompanied by a figure.
You can infere these relationships from the figure
(x-5)° = 74° => x = 74 + 5 = 79°
(x-5)° + 58° + (y-1)° = 180 ° => 74 + 58 + y-1 = 180 =>
y = 180 + 1 - 58 - 74 = 47°
Answer: x = 79°, y = 47°. This is the option d)
I think it is Adjacent and Supplementary
Answer:
In order to find the variance we need to calculate first the second moment given by:
And the variance is given by:
![Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20%2B%5BE%28X%29%5D%5E2%20%3D%2023.36%20-%5B4.74%5D%5E2%20%3D%200.8924)
And the deviation would be:

Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case we have the following distribution given:
X 3 4 5 6
P(X) 0.07 0.4 0.25 0.28
We can calculate the mean with the following formula:

In order to find the variance we need to calculate first the second moment given by:

And the variance is given by:
![Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20%2B%5BE%28X%29%5D%5E2%20%3D%2023.36%20-%5B4.74%5D%5E2%20%3D%200.8924)
And the deviation would be:

Answer= 6.5y+12
distribute the 5 in the parentheses:
(-6y-3) + (15+12.5y)
(you can take away the parentheses after this step)
then add like terms together.
12.5y + (-6y) = 6.5y
15 + (-3) = 12
make it one expression.
6.5y + 12