Answer:
The expectation the instructor to send each day
E(X) = 3.65
Step-by-step explanation:
Given data x : 0 1 2 3 4 5
P(X=x) : 0.05 0.05 0.1 0.1 0.4 0.3
The given data satisfy the two conditions
I) Given all probabilities are p₁(x) ≥0
ii) sum of all probabilities is equal to one
0.05 + 0.05 + 0.1 + 0.1 + 0.4 +0.3 =1
Therefore given data is discrete probability distribution
<u>Expectation</u> :-
suppose a random variable X assumes the values
with respective probabilities
, then the Expectation or Expected value of X , denoted by E(X) = ∑pi xi
![E(X) = 0X0.05+1X 0.05 +2X 0.1 +3X 0.1 +4X0.4 +5X 0.3](https://tex.z-dn.net/?f=E%28X%29%20%3D%200X0.05%2B1X%200.05%20%2B2X%200.1%20%20%2B3X%200.1%20%2B4X0.4%20%2B5X%200.3)
on simplification, we get
E(X) = 3.65
Answer:
$2205
Step-by-step explanation:
equation set up
2000 (1.05)^2
Answer:
1 7 21 35 35 21 7 1
Step-by-step explanation:
Tbh I just searched it up who memorizes this stuff?
B: m is less than or equal to 3
2: 10 pm