The probability distribution for damage claims paid by the newton automobile insurance company on collision insurance follows. payment ($) probability0 0.85500 0.041,000 0.043,000 0.035,000 0.028,000 0.0110,000 0.01(a) use the expected collision payment to determine the collision insurance premium that would enable the company to break even. if required, round your answers to two decimal places. if your answer is zero, enter "0".x f(x) xf(x)0 500 1000 3000 5000 8000 1 total (b) the insurance company charges an annual rate of $520 for the collision coverage. what is the expected value of the collision policy for a policyholder? (hint: it is the expected payments from the company minus the cost of coverage.) if required, enter negative value as negative number. $ 90 why does the policyholder purchase a collision policy with this expected value?

answer: median

step-by-step explanation:

the median is the middle value when a data set is ordered from least to greatest

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1

i would be happy to . most people on would also love to i'm sure. if you just want someone who has done algebra before and want a little more personal relationship than just begging strangers for , i can certainly be of service. i do have college myself so i might not be 24/7.