Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted, with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. Calculate a 90 percent confidence interval for μ, the mean number of unoccupied seats per flight during the past year.

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1

answer: 72: 111

step-by-step explanation:

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)

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.