Answer to your question is 2
6 seashells left is the answer
D=M/V. You just multiply both sides by D to get DV=M, then divide V by both sides.
First you have to make the assumption that these are the only two outcomes. There is also the possibility of hitting the ball and getting out.
However, if we assume that these are the only two cases, we know that the probability is 58.3%. This is because it has been on base 7 times out of 12.
Answer:
A) 68.33%
B) (234, 298)
Step-by-step explanation:
We have that the mean is 266 days (m) and the standard deviation is 16 days (sd), so we are asked:
A. P (250 x < 282)
P ((x1 - m) / sd < x < (x2 - m) / sd)
P ((250 - 266) / 16 < x < (282 - 266) / 16)
P (- 1 < z < 1)
P (z < 1) - P (-1 < z)
If we look in the normal distribution table we have to:
P (-1 < z) = 0.1587
P (z < 1) = 0.8413
replacing
0.8413 - 0.1587 = 0.6833
The percentage of pregnancies last between 250 and 282 days is 68.33%
B. We apply the experimental formula of 68-95-99.7
For middle 95% it is:
(m - 2 * sd, m + 2 * sd)
Thus,
m - 2 * sd <x <m + 2 * sd
we replace
266 - 2 * 16 <x <266 + 2 * 16
234 <x <298
That is, the interval would be (234, 298)