A low-noise transistor for use in computing products is being developed. It is claimed that the mean noise level will be below the 2.5-dB level of products currently in use. It is believed that the noise level is approximately normal with a standard deviation of .8. find 95% CI

The 95% CI is  

Step-by-step explanation:

From the question we are told that

   The  population mean

    The standard deviation is  

Given that the confidence level is  95% then the level of confidence is mathematically evaluated as

   =>  

  =>    

Next we obtain the critical value of   from the normal distribution table, the values is  

Generally the margin of error is mathematically evaluated as

here we would assume that the sample size is  n =  16 since the person that posted the question did not include the sample size

  So    

The  95% CI is mathematically represented as

substituting values

substituting values

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).