Answer:
I believe it's D.
Lmk if I was right, cause I'm 98% sure on this
Altho' I can easily guess what you're supposed to do here, I must point out that you haven't included the instructions for this problem.
I'll help you by example. Let's look at the first problem:
"Evaluate 6(z-1) at z-4."
Due to "order of operations" rules, we must do the work inside the parentheses FIRST. Replace the z inside (z-1) with "-4". We obtain
6(-4-1) = 6(-5) = -30 (answer.)
Your turn. Try the next one. If it's unclear, as questions.
Answer:
$72693.9
Step-by-step explanation:
To get this answer you need to use the compound interest formula, which will be A=P(1+r/n)^n(t). P=59,000 r=11%=0.11 n=1 (annually) t=2 years. From there you should be able to figure the rest out and get the answer. Hope this helps!
Answer: m=
1
3
√2 or m=
−1
3
√2
Step-by-step explanation:
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.
