Out of every 36 children, 9 are wearing a blue shirt.
2 answers:
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Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions </u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
(a)
First, we take its derivative
Solve N'(t)=0
Simplifying
Solving for t
Only t=1 belongs to the valid interval
Taking the second derivative
Which is always positive, so t=1 is a minimum
(b)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
N(15)=4391.7 is a maximum
The standard deviation is 9.27. The typical heart rate for the data set varies from the mean by an average of 9.27 beats per minute.
<h3>How to determine the standard deviation of the data set?</h3>The dataset is given as:
Heart Rate Frequency
60 1
65 3
70 4
75 12
80 8
85 15
90 9
95 5
100 3
Calculate the mean using
Mean = Sum/Count
So, we have
Mean = (60 * 1 + 65 * 3 + 70 * 4 + 75 * 12 + 80 * 8 + 85 * 15 + 90 * 9 + 95 * 5 + 100 * 3)/(1 + 3 + 4 + 12 + 8 + 15 + 9 + 5 + 3)
Evaluate
Mean = 82.25
The standard deviation is
So, we have:
SD = √[1 * (60 - 82.25)^2 + 3 * (65 - 82.25)^2 + 4 * (70 - 82.25)^2 + 12 * (75 - 82.25)^2 + 8 * (80 - 82.25)^2 + 15 * (85 - 82.25)^2 + 9 * (90 - 82.25)^2 + 5 * (95 - 82.25)^2 + 3 * (100 - 82.25)^2)]/[(1 + 3 + 4 + 12 + 8 + 15 + 9 + 5 + 3 - 1)]
This gives
SD = √85.9533898305
Evaluate
SD = 9.27
Hence. the standard deviation is 9.27. The typical heart rate for the data set varies from the mean by an average of 9.27 beats per minute.
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