Answer:
I think 47 and 55
Explanation:
I think that each number is the previous one plus 8;
7+8=15
15+8=23
23+8=31
31+8=39
So, next will be:
39+8=47
47+8=55
Is this what ur looking for?? hope this help:)
Answer:
a) 81.5%
b) 95%
c) 75%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 266 days
Standard Deviation, σ = 15 days
We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.
Formula:

a) P(between 236 and 281 days)

b) a) P(last between 236 and 296)

c) If the data is not normally distributed.
Then, according to Chebyshev's theorem, at least
data lies within k standard deviation of mean.
For k = 2

Atleast 75% of data lies within two standard deviation for a non normal data.
Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.
Answer: 230.697
Step-by-step explanation:
Answer:
yes
Step-by-step explanation:
Answer:
47
Step-by-step explanation:
since each gallon is 4 quarts you would multiply by 11. 44 quarts then add 3 extra quarts to get 47