Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that
P(Ch) = 88/520
P(C) = 76/520
P(Ch and C) = 31/520
and we want to find P(Ch or C).
Using the inclusion/exclusion principle, we have
P(Ch or C) = P(Ch) + P(C) - P(Ch and C)
P(Ch or C) = 88/520 + 76/520 - 31/520
P(Ch or C) = 133/520
Answer:
Length = 5
Width = 21
Step-by-step explanation:
(x)(x + 16) = 105
x^2 + 16x = 105
x^2 + 16x - 105 = 0
(x - 5) x ( x + 21) = 0
x - 10 = 0
x = 5
x + 21 = 0
x = -21
Now that we have the zeroes.
We have to find the most viable one to put in.
Using -21 would not make sense, so we will use 5.
Plug it in:
x = 5
(5) (5 + 16) = 105
5 ( 21) = 105
2X^3-2x-40=0
Hope this helps
Answer:
15-16=-1
Step-by-step explanation:
