PLEASE HEL A standard I. Q. test produces normally distributed results with a mean of 100 and a standard deviation of 15 for the city of New York. Out of approximately 8,400,000 citizens, how many of these people would have I. Q.s below 67?

approx  193200

Step-by-step explanation:

As known for normal distribution is correct the rule 95.4% of the results are situation within mean+-2*s  ( where s is a standard deviation)

So the border is 100+-2*15=70 and that is approx=67.

95.4% of 84000000 citizens are= 8 400 000*0.954=8013600 persons

So the residual number of the citizens =8400000-8013600=386400 citizens

Because of the simmetry of normal distribution to find the number of the citizens that have IQ below 67 we have to divide 386400 by 2.

N=386000/2=193200

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

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

10 - 75

step-by-step explanation:

there's nothing attached so it's most likely anything in that range.