Answer and Step-by-step explanation:
First, we need to find out the circumference of the circle.
We know that the circle has a radius of 1, and that we are finding the circumference. We'll use the circumference equation of a circle.
2
r
<u>Plug in 1 for r.</u>
=
= circumference of circle
<u>Now, multiply the answer (the circumference) by 4 to get the perimeter of the square.</u>
<u />
x 4 =
= perimeter of square
<u>Now, divide the perimeter by 4 to get what the side of the square is.</u>
<u />
÷ 4 = 
<u>Now, multiply </u>
<u> by </u>
<u> (side times side) to get the area.</u>
x
= 
The answer is A. 
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
<em><u>I hope this helps!</u></em>
<em><u>Brainliest is appreciated,</u></em>
The standard form of a circle is:

Your center is
and your radius is 
Plug them into the equation and you get


The correct answer is D.
True. When the points are plotted on a graph and connected, they pass the vertical line test. No x value is the same.
Answer:
Step-by-step explanation:
yard; the ratio of feet to yards is
Answer:
a)
b) 
c) 
d) 
Step-by-step explanation:
Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that
The probability mass function for the random variable is given by:
And f(x)=0 for other case.
For this distribution the expected value is the same parameter
,
, 
a. Compute both P(X≤4) and P(X<4).
Using the pmf we can find the individual probabilities like this:




b. Compute P(4≤X≤ 8).







c. Compute P(8≤ X).


d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?
The mean is 4 and the deviation is 2, so we want this probability




