Answer:
Step-by-step explanation:
To find (c.g)(x) we basically use the function d(x) as your variable, x, and plug it into c(x). So we replace every x in c(x) with the function d(x),
c(g(x))=4(x^{2} +5x)-2
c(g(x))4+20x-2
The time for a professor to grade a student's homework in statistics is normally distributed with a mean of 12.6 minutes and a s
1 answer:
Answer:
37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that randomly selected homework will require between 8 and 12 minutes to grade?
This is the pvalue of Z when X = 12 subtracted by the pvalue of Z when X = 8. So
X = 12
has a pvalue of 0.4052
X = 8
has a pvalue of 0.0329
0.4052 - 0.0329 = 0.3723
37.23% probability that randomly selected homework will require between 8 and 12 minutes to grade
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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.