Answer:
Answer:
They both have q+3/2p, so that means that 2PQ=CB and that means they are parallel to each other
Step-by-step explanation:
PQ=PA+QA
PQ=1/2(2q-p)+2/5*5p=q-1/2p+2p=q+3/2p
CB=2q+3p=2(q+3/2p)
Other explanation: It should be written like this PQ=q+3/2p and CB=2q+3p=2(q+3/2p) they are parallel bcs CB=2*PQ.
Since you're looking for the chance that the defective player occurs twice, you need to find the chance your friend receives a defective player given that you also receive one. The chance you receive a defective player is 4%, or 0.04. If you friend also receives a defective player, then the chance of both occurring is 4% of 4%, or 0.04 * 0.04, which equals 0.0016. So the probability that you can a friend both receive a defective player is 0.16%.
That's a bit of a nasty. Where ever you see (x^2 + 6) you put in p.
p^2 - 21 = 4(x^2 + 6)
p^2 - 21 = 4p
p^2 - 4p - 21 = 0
A <<<< answer.
![y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\ y''(0)=20\cdot0^3=0](https://tex.z-dn.net/?f=y%3Dx%5E5-3%5C%5C%20y%27%3D5x%5E4%5C%5C%5C%5C%205x%5E4%3D0%5C%5C%20x%3D0%5C%5C%200%5Cin%20%5B-2%2C1%5D%5C%5C%5C%5C%20y%27%27%3D20x%5E3%5C%5C%5C%5C%0Ay%27%27%280%29%3D20%5Ccdot0%5E3%3D0)
The value of the second derivative for
is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of
is always positive for
. That means at
there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval
![[-2,1]](https://tex.z-dn.net/?f=%5B-2%2C1%5D)
.
The function
is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.
