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)
A. -14x+21
because -7 times 2x is -14x and -7 times -3 is 21
Answer:
y=350x+125
Step-by-step explanation:
y=350x+125
Answer:
0.8 or .8
Step-by-step explanation:
multiply the top and bottom by 10
so 8/10=80/100
to get the decimal form, just divide 8 by 10 and it should give you 0.8
