There would be a 10% chance that they would have the same birthday
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:
12.5 pi or 39.26990
Step-by-step explanation:
First we can find the radius, which is half the diameter, 5 feet. To find the area of a semi-circle we would find the area of the circle and then divide by 2:
We can substitute what we have:
This gives us:
12.5 pi or 39.26990(Not sure what format you need it in)
200 - (2x) = 42
200 - 42 = 2x
158/2 = 2x/2
x = 79
he is 79 years old
hope this helps
a wise 'old' man
Answer:
395/136
Step-by-step explanation:
(9 7/8)/(3 2/5) = (79/8)/(17/5) = (79/8)(5/17) = 395/136
Since 9 7/8 is larger than 3 2/5, it is an improper fraction.
