A copper smelting process is supposed to reduce the arsenic content of the copper to less than 1000 ppm. Let μ denote the mean arsenic content for copper treated by this process, and assume that the standard deviation of arsenic content is σ = 140 ppm. The sample mean arsenic content X¯¯¯ of 95 copper specimens will be computed, and the null hypothesis H0: μ ≥ 1030 will be tested against the alternate H1: μ < 1030. What is the power of a 10% level test if the true mean content is 985 ppm? Round the answer to four decimal places.

The power is

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

answer: the answer is ann

the answer is b.

step-by-step explanation: