We know that 30 students have 17 dogs. If we want to know how many dogs will have 10 times more students (300) we can miltiply 17 times 10 as well to get the answer.
17*10=170 - its predicted value
Differentiate it
get f ' (x)=3x^2-4
put 3x^2-4=0
x=2/√3,-2/√3
so interval
x∈(-2/√3,2/√3)
Answer:
Step-by-step explanation:
let x and y be length and width of rectangle.
Perimeter=2(x+y)
area=xy
2(x+y)=1/2 xy
4(x+y)=xy
4x+4y=xy
x y-4y=4 x
y(x-4)=4 x
Answer: 0.02
Step-by-step explanation:
OpenStudy (judygreeneyes):
Hi - If you are working on this kind of problem, you probably know the formula for the probability of a union of two events. Let's call working part time Event A, and let's call working 5 days a week Event B. Let's look at the information we are given. We are told that 14 people work part time, so that is P(A) = 14/100 - 0.14 . We are told that 80 employees work 5 days a week, so P(B) = 80/100 = .80 . We are given the union (there are 92 employees who work either one or the other), which is the union, P(A U B) = 92/100 = .92 .. The question is asking for the probability of someone working both part time and fll time, which is the intersection of events A and B, or P(A and B). If you recall the formula for the probability of the union, it is
P(A U B) = P(A) +P(B) - P(A and B).
The problem has given us each of these pieces except the intersection, so we can solve for it,
If you plug in P(A U B) = 0.92 and P(A) = 0.14, and P(B) = 0.80, you can solve for P(A and B), which will give you the answer.
I hope this helps you.
Credit: https://questioncove.com/updates/5734d282e4b06d54e1496ac8
Answer:
Step-by-step explanation:
So we are assuming only the times listed count toward telling us how long the game was, so you just add the three numbers together. 155, 175 and 30
