The mean amount spent by a family of four on food per month is $500 with a standard deviation of $75. assuming a normal distribution, what is the probability that a family spends more than $410 per month? a. 0.1151 b. 0.1539 c. 0.8849. d. 0.8869 e. none of the above

Let X be the amount spent by a family of four on food per month.

The mean amount spent per month on food is μ = 500

The standard deviation σ =75

X follows Normal distribution with mean μ = 500 and standard deviation σ =75.

The probability that a family spends more than $410 per month is

P(x > 410) =

= P(z > -1.2)

= 1 - P(z < -1.2)

Using z score to find probability below -1.2

= 1 - 0.1151

P(x > 410) = 0.8849

The probability that a family spends more than $410 per month is 0.8849

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: