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Mathematics

1 answer:

Wewaii [24]2 years ago

8 0

Answer:

36/25

Step-by-step explanation:

Hope it helps you!

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How to solve the 4a.plz

vladimir1956 [14]

(4xy-2y^2)+2y
4-2=2....x will remain y^1-y^2=y^-1
2xy^-1+2y
answer=2x+2y

3 0

2 years ago

Suppose that p is the probability that a randomly selected person is left handed. The value (1-p) is the probability that the pe

solmaris [256]

Answer:

a) 1/2

b) 250

Step-by-step explanation:

The start of the question doesn't matter entirely, although is interesting to read. What we are trying to do is find the value for $p$ such that $1000p(1-p)$ is maximized. Once we have that $p$ , we can easily find the answer to part b.

Finding the value that maximizes $1000p(1-p)$ is the same as finding the value that maximizes $p(1-p)$ , just on a smaller scale. So, we really want to maximize $p(1-p)$ . To do this, we will do a trick called completing the square.

$p(1-p)=p-p^2=-p^2+p=-(p^2-p)=-(p^2-p+1/4)-(-1/4)=-(p-1/2)^2+1/4$ .

Because there is a negative sign in front of the big squared term, combined with the fact that a square is always positive, means we need to find the value of $p$ such that the inner part of the square term is equal to $0$ .