B. 5.0 minutes
C. 5.4 minutes
D. 5.8 minutes
E. 6.0 minutes
Set B: 5, 8, 7, 6, 4
Mean of Set A is 5 and Set B is 6. Standard deviation of Set A is approximately 2.83, and for Set B, it's approximately 1.67. This indicates that values in Set B are generally closer to their mean than values in Set A to their mean.
To compare the mean and standard deviation of Set A and Set B, we first need to calculate these for each set. Mean is the average of the numbers and standard deviation is a measure of the amount of variation or dispersion of a set of values.
First, calculate the mean by adding the numbers in each set and dividing by the total number of values. For Set A, the mean is (7+3+4+9+2)/5 = 5. For Set B, the mean is (5+8+7+6+4)/5 = 6.
The standard deviation is a bit more complex, as it involves subtracting the mean from each value, squaring the result, finding the mean of these squares, and then taking the square root of that mean. For Set A, these steps result in a standard deviation of approximately 2.83. For Set B, these steps result in a standard deviation of approximately 1.67.
In conclusion, Set B has a higher mean and a lower standard deviation compared to Set A which means values in Set B are generally closer to the mean of Set B than values in Set A are to the mean of Set A.
#SPJ12
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Work out the value of v when
u = 10 and t = 7
if u=10 and t=7
first: plug in the numbers (V=10+10(7))
second: use PEMDAS to solve
Third: 10+ 70=80
V=80
8/13 and 7/10.
3/5 − 6/11
A:3/55
B:3/6
C:3/11