Answer:
The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018
Step-by-step explanation:
The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.
The cummulative distribution function of W will be denoted by
. The values of
can be found in the attached file.

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.
1.
Answer B
2.
Answer D.
3.
Answer D.
4.
Answer C.
Answer:
D
Step-by-step explanation:
Given
x² - 7x + 12
Consider the factors of the constant term (+ 12) which sum to give the coefficient of the x- term (- 7)
The factors are - 4 and - 3, since
- 4 × - 3 = + 12 and - 4 - 3 = - 7, thus
x² - 7x + 12 = (x - 4)(x - 3) ← in factored form → D