Answer:
(b) 1/792
Step-by-step explanation:
The complete question is;
<em>Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.</em>
<em>(a) How many different groups of 5 problems can be chosen from the 12</em>
<em>problems?</em>
<em>(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?</em>
<u>In (a)</u>
<u></u>
Apply the formula

where n=12 and r=5
substitute values

In (b)
If Jerry did only 5 problems of one assignment then the probability will be

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The graph is going to be all three, because when you throw the ball it goes higher, then constant, then comes back down. It has a maximum point because you can only throw the ball so high up, then it starts to come back down.
If this helped then please give brainliest answer, rate, and thank!
Answer:
A cross section parallel to the base is a square measuring 4 cm by 4 cm.
A cross section perpendicular to the base through the midpoints of opposite sides is a square measuring 4 cm by 4 cm.
A cross section that passes through the entire bottom front edge and the entire top back edge is a rectangle measuring 4 cm by greater than 4 cm.
Step-by-step explanation:
Cross sections parallel and Perpendicular to the base are squares 4 × 4
The diagonal cross section will be a rectangle with base 4 and height
sqrt(4² + 4²) = 4sqrt(2) > 4
Answer:


![V(X) = E(X^2)-[E(X)]^2=349.2-(18.6)^2=3.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%28X%5E2%29-%5BE%28X%29%5D%5E2%3D349.2-%2818.6%29%5E2%3D3.24)
The expected price paid by the next customer to buy a freezer is $466
Step-by-step explanation:
From the information given we know the probability mass function (pmf) of random variable X.

<em>Point a:</em>
- The Expected value or the mean value of X with set of possible values D, denoted by <em>E(X)</em> or <em>μ </em>is

Therefore

- If the random variable X has a set of possible values D and a probability mass function, then the expected value of any function h(X), denoted by <em>E[h(X)]</em> is computed by
![E[h(X)] = $\sum_{D} h(x)\cdot p(x)](https://tex.z-dn.net/?f=E%5Bh%28X%29%5D%20%3D%20%24%5Csum_%7BD%7D%20h%28x%29%5Ccdot%20p%28x%29)
So
and
![E[h(X)] = $\sum_{D} h(x)\cdot p(x)\\E[X^2]=$\sum_{D}x^2\cdot p(x)\\ E(X^2)=16^2\cdot 0.3+18^2\cdot 0.1+20^2\cdot 0.6\\E(X^2)=349.2](https://tex.z-dn.net/?f=E%5Bh%28X%29%5D%20%3D%20%24%5Csum_%7BD%7D%20h%28x%29%5Ccdot%20p%28x%29%5C%5CE%5BX%5E2%5D%3D%24%5Csum_%7BD%7Dx%5E2%5Ccdot%20p%28x%29%5C%5C%20E%28X%5E2%29%3D16%5E2%5Ccdot%200.3%2B18%5E2%5Ccdot%200.1%2B20%5E2%5Ccdot%200.6%5C%5CE%28X%5E2%29%3D349.2)
- The variance of X, denoted by V(X), is
![V(X) = $\sum_{D}E[(X-\mu)^2]=E(X^2)-[E(X)]^2](https://tex.z-dn.net/?f=V%28X%29%20%3D%20%24%5Csum_%7BD%7DE%5B%28X-%5Cmu%29%5E2%5D%3DE%28X%5E2%29-%5BE%28X%29%5D%5E2)
Therefore
![V(X) = E(X^2)-[E(X)]^2\\V(X)=349.2-(18.6)^2\\V(X)=3.24](https://tex.z-dn.net/?f=V%28X%29%20%3D%20E%28X%5E2%29-%5BE%28X%29%5D%5E2%5C%5CV%28X%29%3D349.2-%2818.6%29%5E2%5C%5CV%28X%29%3D3.24)
<em>Point b:</em>
We know that the price of a freezer having capacity X is 60X − 650, to find the expected price paid by the next customer to buy a freezer you need to:
From the rules of expected value this proposition is true:
We have a = 60, b = -650, and <em>E(X)</em> = 18.6. Therefore
The expected price paid by the next customer is

Answer:
112°
Step-by-step explanation:
Given that the diagonals of trapezoid RSTU are congruent, it is an Isosceles Trapezoid.
One of the properties of an Isosceles Trapezoid is that the base angles are equal.
Therefore if the measure of angle S=112°, the measure of Angle U will also be 112°.