A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 449.0 gram setting. It is believed that the machine is underfilling the bags. A 37 bag sample had a mean of 441.0 grams. A level of significance of 0.05 will be used. Determine the decision rule. Assume the standard deviation is known to be 18.0.

the decision rule is to reject H0 if Z is < -1.645

Step-by-step explanation:

this question is interested in the decision rule.

H0: u = 449

H1 : u <449

We find the test statistic

Mean bar x = 441

u = 449

Sd = 18

n = 37

Z = barX - u / sd/√n

= 441-449/18/√37

= -8/18/6.083

= -8/2.959

Z = - 2.7036

This is the value of the test statistic

The critical value at 0.05 = -1.645. this is a left tailed test

The decision rule is to reject H0 if Z < -1.645

The test statistic is less than the critical value

-2.7036 < -1.645

So we have enough evidence to reject H0. At 0.05 level we have enough evidence to support claim.

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.

step-by-step explanation:

it will be a cylinder. with a diameter of 8 and 4 single pieces high