I’m questioning whether or not the Gacha life thing matters or not.
For 27) D
For 28) A
Hope that helps!
Hopefully this makes sense!
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:
3.6 hours or 3 hours and 36 minutes.
Step-by-step explanation:
If the first pipe can fill the pool in 9 hours that means that in 1 hour it fills of the pool, we use the same logic for the second pipe and conclude that in 1 hour it feels up of the pool. If both were to fill the pool together that means that in 1 hour they would fill...
of the pool in 1 hour.
Now to find how many hours it will take divide the pool by the speed with which it is getting filled.
hours
3.6 hours in other words is 3 hours and 36 minutes.