According to the college board website, the scores on the math part of the sat (sat-m) in a recent year had a mean of 507 and standard deviation of 111. assume that sat-m scores follow a normal distribution. one of the criteria for admission to a certain engineering school is an sat-m score in the 98th percentile. this means the score is in the top 2% of scores. how does this translate to an actual sat-m score?

the answer is 40.

proof:

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81

first the equation needs to be simplified. so the equation would now look like:   52-24+11

then you multiply.

and when you multiply, you will get the answer of 39.