Answer:

Step-by-step explanation:
Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.
-We calculate the probability of a mean of 3.4 as follows:
#First determine the z-value:

#We then determine the corresponding probability on the z tables:

Hence, the probability of obtaining a sample mean this large or larger is 0.0228
The rate a computer works is 1/time. Working together, you add the rates.
Let new computer be x, old computer be y.
x = y - 7

Rounded to nearest tenth gives:
x = 23 hours
Answer:
The last two functions are decreasing!
Step-by-step explanation:
You can easily tell the functions are decreasing if the function is going down from L to R. If it is going up, from L to R, then it is increasing!
The school year would be 1981-1982