A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation 0.25 volt, and the manufacturer wishes to test volts against volts, using units. In your intermediate calculations, use z-scores rounded to two decimal places (e. g. 98.76). (a) The acceptance region is. Find the value of a.
(b) Find the power of the test for detecting a true mean output voltage of 5.1 volts.

hello your question was poorly written but i was able to the get missing parts online which enabled me resolve your question

a) a = 0.1096

b) 1.89 watts

Step-by-step explanation:

Std of output voltage = 0.25 volt

H0 : μ = 5 volts

Ha : μ ≠ 5 volts

n = 16

a) Acceptance region = 4.9 ≤ X ≤ 5.1  

Determine the value of a

value of a = 0.0548 + 0.0548

                 = 0.1096

attached below is the reaming solution

note : a is a type 1 error

b) power of test

True mean output voltage = 5.1 volts

P = - 1.89 watts

power cant be negative hence the power of the test  = 1.89 watts

the circle graph is not a function.

step-by-step explanation:

the graph of the circle is not a function.

because for a graph to be a function, each value of x must give one value of y

in the circle, the values of x for -2< x< 2 each give 2 values of y.

hence this is not a function.

the correct answers i think are b. and d.

give me brainliest if it is correct!