Answer: 1/36
Hope this helps. - M
Hewo the whole world that i hate^^
2 answers:
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Answer: 1/36
Hope this helps. - M
Answer:
The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.
Step-by-step explanation:
The mean sample of the sum of n random variables is
If are indentically distributed and independent, like in the situation of the problem, then the variance of
will be the sum of the variances, in other words, it will be n times the variance of
.
However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus and as a result, the standard deviation of
is the standard deviation of
divided by
.
Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:
- Restaurants: We have boxes with 9 salmon each, so it will be
- Grocery stores: Each carton has 49 salmon, thus the standard deviation is
- Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is
We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.
Answer:
Step-by-step explanation:
Y) = 3
*multiply both sides by 8 - cancels out 8 in denominator*
x + 4 = 24
*subtract 4 from both sides*
x = 20
E) = 1
*multiply both sides by 2 - cancels out 2 in denominator*
x - 5 = 2
*add 5 on both sides*
x = 7
N) = 2
* multiply both sides by 4 - cancels out 4 in denominator*
x + 2 = 8
*subtract 2 from both sides*
x = 6