To problem wants to compute the probability of the expected value that could be earn if you would draw again another card. So base on the fact that if draw a face card of K,Q,J from a standard deck correctly you will earn 10 point in just one draw, but loose 2 points if you another card, the probability that you would draw correct card in just one draw is 30.77%, so the possible value earn you have 2 points only
GIRL- THIS HAPPENS TO ME TOO I TJINK YOU POOPED ACCIDENTALLY
Answer:
(2x-3) (2x+3)
zeros, x intercepts: -3/2, 3/2
Step-by-step explanation:
4x^2 -9
We know the difference of squares is a^2 -b^2
This factors into (a-b) (a+b)
Let 4x^2 =a^2
Taking the square root
2x =a
Let b^2 =9
Taking the square root
b= 3
(4x^2-9 ) = (2x-3) (2x+3)
To find the zeros, we set the equation equal to zero
(4x^2-9 ) = (2x-3) (2x+3) =0
Using the zero product property
2x-3 =0 and 2x+3 =0
2x-3+3 = 0+3 2x+3-3 = 0-3
2x=3 2x=-3
Divide by 2
2x/2 = 3/2 2x/2 = -3/2
x = 3/2 x = -3/2
These are the zeros of the equation (which are also the x intercepts)
Answer:
It's false see 4 cannot go into 30 can I have
brainly?:)
Step-by-step explanation:
