The expected value of the discrete distribution, if you have to pay $.50 to pick one package at random, is of -$0.08.
<h3>What is the mean of a discrete distribution?</h3>
The expected value of a discrete distribution is given by the <u>sum of each outcome multiplied by it's respective probability</u>.
For this problem, considering the cost of $0.5, the distribution is given as follows:
- P(X = 0.2) = 12/(12 + 15 + 23) = 12/50 = 0.24.
- P(X = -0.1) = 15/(12 + 15 + 23) = 15/50 = 0.3.
- P(X = -0.2) = 23/(12 + 15 + 23) = 23/50 = 0.46.
Hence the expected value is given by:
E(X) = 0.2 x 0.24 - 0.3 x 0.1 - 0.2 x 0.46 = -$0.08.
More can be learned about the expected value of a discrete distribution at brainly.com/question/13008984
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1 pint = 20 fl. oz
20/5 = 4 cups can be filler per pint
4*2.5 = 10 cups can be filled overall
Hey! We can’t see a picture of the problem.
Answer:
(f + g)(x) = 3x^2 + 3x/2 - 9
Step-by-step explanation:
In order to find a composite function through addition, you simply add the two equations together.
f(x) + g(x)
x/2 - 3 + 3x^2 + x - 6
3x^2 + 3x/2 - 9
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
