Other expressions for 19 times 25 + 19 times 75
1 answer:
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Answer:
a) 151lb.
b) 6.25 lb
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
In this problem, we have that:
So
a) The expected value of the sample mean of the weights is 151 lb.
(b) What is the standard deviation of the sampling distribution of the sample mean weight?
This is
Explanation![\begin{gathered} f(19)=\frac{3}{19+2}-\sqrt[]{19-3}, \\ \\ f(19)=\frac{3}{21}-\sqrt[]{16}, \\ \\ f(19)=\frac{1}{7}-4, \\ \\ f(19)=\frac{1-28}{7}, \\ \\ f(19)=-\frac{27}{7}\text{.} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20f%2819%29%3D%5Cfrac%7B3%7D%7B19%2B2%7D-%5Csqrt%5B%5D%7B19-3%7D%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D%5Cfrac%7B3%7D%7B21%7D-%5Csqrt%5B%5D%7B16%7D%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D%5Cfrac%7B1%7D%7B7%7D-4%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D%5Cfrac%7B1-28%7D%7B7%7D%2C%20%5C%5C%20%20%5C%5C%20f%2819%29%3D-%5Cfrac%7B27%7D%7B7%7D%5Ctext%7B.%7D%20%5Cend%7Bgathered%7D)
Answer
This exercise is about evaluating a function at a particular argument. To do that, we replace the variable with the argument in the formula of the function, and simplify.
Let's do that: