A manufacturer makes ball bearings that have a target diameter of 5\text{ mm}5 mm5, start text, space, m, m, end text. Quality control engineers sample ball bearings from a batch to test the hypotheses H_0: \mu = 5\text{ mm}H 0 :μ=5 mmH, start subscript, 0, end subscript, colon, mu, equals, 5, start text, space, m, m, end text vs. H_\text{a}: \mu \neq 5\text{ mm}H a :μ =5 mmH, start subscript, start text, a, end text, end subscript, colon, mu, does not equal, 5, start text, space, m, m, end text, where \muμmu is the true mean diameter of the ball bearings in that batch. Suppose that H_0H 0 H, start subscript, 0, end subscript is actually true. Which situation below would have the lowest probability of a Type I error?

answer: h

step-by-step explanation:

i think it is 18

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of