Answer:
It is impossible to simulate this with a random number generator without knowing what the correct answer choices were
Step-by-step explanation:
Given
Required
How can he select the right answer
Using randint will only generate a random number which could or could not be the answer to the question.
This is so because each of the 5 options for the question has the same probability of 1/5. So, using randint will only generate a random number. This generated random number has 1/5 chance of being the answer and 4/5 of not being the answer to the question.
In a nutshell, he can not make use of a simulator to select the answer to the questions in this scenario, unless he knows the solution.
<em>Hence (a) answers the question.</em>
Answer:
The cutoff sales level is 10.7436 millions of dollars
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:
15th percentile:
X when Z has a pvalue of 0.15. So X when Z = -1.047.
The cutoff sales level is 10.7436 millions of dollars
New cordinates are formed by adding 7 in x and subtracting 2 from y
A(−2, 2) =A ' (-2 +7 , 2 - 1 ) = A' (5,1)
B(−2, 4) = B' (-2 + 7 , 4 -1 )= B' (5,3)
C(2, 4) = C' (2 + 7 , 4 -1 )= C' (9,3)
<span>D(2, 2) D' (2 + 7 , 2 -1 ) = D' ( 9 , 1)</span><span>
</span>
Area of Shaded triangle as a whole: 136cm^2
Area of Rectangle 1 : 20cm^2
Area of Rectangle 2 : 6cm ^ 2
Area of the shaded region :
136 - (20+6)
136 - (26)
110 cm^2
Answer:
The standard error of the mean is 4.5.
Step-by-step explanation:
As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:
The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.
The mean of the sample is:
The standard deviation of the sample is calculated as:
Then, we can calculate the standard error of the mean as:
