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vaieri [72.5K]
3 years ago
9

Urgent please help..

Mathematics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

5.4

Step-by-step explanation:

First find the area of the square.

L times W

5 times 5=25

Now find the area of the cricle

A=\pir^{2}

Substitute pi with 3.14 and radius with 2.5

A= 3.14 times2.5^{2}

A=3.14 times 6.25

A=19.625

Now substract the area of the square from the area of the circle

Your answer will be 5.375

The rounded answer will be 5.4

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Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a
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Answer:

Mean = 68.9

s^2 =18.1 --- Variance

Step-by-step explanation:

Given

\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5

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x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5

For 67 - 74

x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5

For 75 - 82

x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5

For 83 - 90

x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5

So, the table becomes:

\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

The mean is then calculated as:

Mean = \frac{\sum fx}{\sum f}

Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}

Mean = \frac{2547.5}{37}

Mean = 68.9 -- approximated

Solving (b): The sample variance:

This is calculated as:

s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}

So, we have:

s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}

s^2 =\frac{652.8}{36}

s^2 =18.1 -- approximated

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