Answer:
About 99.7% of births would be expected to occur within 51 days of the mean pregnancy length
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Standard deviation = 17.
About what percentage of births would be expected to occur within 51 days of the mean pregnancy length?
51/17 = 3.
So, within 3 standard deviations of the mean.
About 99.7% of births would be expected to occur within 51 days of the mean pregnancy length
Answer:what’s the lowest value?
what’s the highest value?
there is your range
Step-by-step explanation:
If we know that Pedro bought 1

pounds of red sand, and 5

pounds of brown sand, and they both cost $4.60, then we just have to add the individual amounts of sand, since they cost the same.

So Pedro bought

lbs of sand overall. Now, we just need to multiply the cost per pound by the total amount of sand he bought.

Pedro spent $31.05 on sand, which is answer choice B.
Hope that helped =)
Answer: 3.712 hours or more
Step-by-step explanation:
Let X be the random variable that denotes the time required to complete a product.
X is normally distributed.

Let x be the times it takes to complete a random unit in order to be in the top 10% (right tail) of the time distribution.
Then, 
![P(z>\dfrac{x-3.2}{\sigma})=0.10\ \ \ [z=\dfrac{x-\mu}{\sigma}]](https://tex.z-dn.net/?f=P%28z%3E%5Cdfrac%7Bx-3.2%7D%7B%5Csigma%7D%29%3D0.10%5C%20%5C%20%5C%20%5Bz%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D)
As,
[By z-table]
Then,

So, it will take 3.712 hours or more to complete a random unit in order to be in the top 10% (right tail) of the time distribution.
Answer: I think your answer would be B
i hope it's right