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:
the anwser is b
Step-by-step explanation:
divide 27 by 180 and you get an answer of 54
F(x)= (x+4) ^{2} -13 is the answer
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Answer:
(x+3)²-11=0
Step-by-step explanation:
0=9(x2+6x)-18
0= 9x² +54x-18
0= 9( x²+6x-2) taking 9 common
0= x² +6x-2 (breaking the mid term to find the second element of the square
0 = (x)² +2(x)(3) +(3)² -2 -9 adding and subtracting the second element
0= (x+3)² -11 summing up
Answer:
10 units ^2
Step-by-step explanation:
Area 1:
A = lw
= 2 * 5 = 10
