the three numbers are 49,56 and 63.
Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
Answer:
-4
Step-by-step explanation:
-4(3) = -12
-12 + 8 = -4
Answer:
-10
Step-by-step explanation:
as the -32 is in bracket so we must look at it first. then the (-) is multiplied with the + outside bracket which is +×(-)=- . so it will be plus i.e. 22-32 and final step subtract it. the answer is -10.
