Answer:
Step-by-step explanation:
The Answer is {x| x <-4 or x> 2}
Q(p) = k/p^3 . . . . . . . . we want to find k
q(10) = k/10^3 = 64
k = 64,000
Revenue = q(p)*p = 64000/p^2
Cost = 150 +2q = 150 +2*64000/p^3
Profit = Revenue -Cost = 64000/p^2(1 -2/p) -150
Differentiating to find the maximum profit, we have
.. dProfit/dp = -2(64000/p^3) +6(64000/p^4) = 0
.. -1 +3/p = 0
.. p = 3
A price of $3 per unit will yield a maximum profit.
So,
15/390 = 5/130 = 1/26
In decimal form, that is 0.03846153846...
This problem is easier solved by finding the probability that she does NOT do her homework both Monday and Tuesday, which is obtained by the multiplication rule.
P(no HW on Monday) = 1-0.75 = 0.25
P(no HW on Tuesday) = 1-0.75 = 0.25
P(no HW on both Monday and Tuesday) = 0.25*0.25=0.0625
[by the multiplication rule]
This means that the rest of the time (1-0.0625=0.9375) Elsie does her homework either Monday, or Tuesday, or both days.
=>
P(HW either Monday, Tuesday, or both) = 0.9375
(note: in current English, Monday or Tuesday means "either Monday, Tuesday, or both days")
It would be all the evens since the reference is x^2... it’s not the odds Bc it’s not a graph like or
So jt would be 4,6,8
Hope it helps
Ps I’m not positive, haven’t done this for a while now