Aset of score has a mean of μ = 63 and a standard deviation of σ = 8. if these scores are standardized so that the new distribution has μ = 50 and σ = 10, what new value would be obtained for a score of x = 59 from the original distribution?

The z-score of 59 in the original distribution is

z = (59 - 63)/8 = -1/2

The corresponding x-score in the new distribution is

-1/2 = (x - 50)/10

-5 = x - 50

45 = x

The new value of 45 would be obtained for a score of 59 from the original distribution.

828. all there is to it!

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10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )