How do i solve 2(x+4)=5x-2

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

spin [16.1K]

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.

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

For 59 - 66

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

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

5 0

3 years ago

I really need help with this answer can someone help?

valina [46]

Answer:

$7075 or 7718

Step-by-step explanation:

91100-6200=84900

84900/11= 7718.18181818 or rounded= 7718

(This is if the december month doesnt count.)

84900/12=7075

(If december is included.)

7 0

3 years ago

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A business owner collected data on the people working for her company. She determined the average number of minutes each employe

Naddika [18.5K]

Answer:

The comparison using median and IQR is best because one of the graphs is not symmetrical.

Step-by-step explanation:

The following information is missing

A box plot titled Number of Minutes Women Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 54, and the box ranges from 34 to 50. A line divides the box at 48.

A box plot titled Number of Minutes Men Spend on Breaks. The number line goes from 30 to 70. The whiskers range from 30 to 68, and the box ranges from 36 to 60. A line divides the box at 48.

The business owner uses the median and IQR to determine the center and variability of the data sets. Which best describes the comparison?

The comparison would be more accurate using the mean and MAD because one of the graphs is symmetric.

The comparison would be more accurate using the mean and MAD because the median of both data sets is the same.

The comparison using median and IQR is best because one of the graphs is not symmetrical.

The comparison using median and IQR is best because the median is greater than the IQR for both data sets.

Mean and MAD are useful for comparison when both data sets are symmetrical.

In the women box plot the Q1 is at 34, the median is at 48, and the Q3 is at 50, so it is not symmetrical (the difference between the median and the Q1, and the Q3 and the median is not the same)

8 0

4 years ago

2 × 10^12 red blood cells have a total mass of 90 grams.

galina1969 [7]

Answer:

2×10¹² cells → 90 g

mass of 1 red blood cell : 90÷(2×10¹²) = 45×10^-12 g = 4.5×10^-11 g

5 0

3 years ago

Pls help BYU question Geometery Transversal,parallel lines

Rufina [12.5K]

Answer:

<18 = 40

Step-by-step explanation:

<3 = <19 since they are alternate exterior angles and alternate exterior angles are equal for parallel lines cut by a transversal

<19 = 140

<19+<18 = 180 since they form a straight line

140 + <18 = 180

Subtract 140 from each side

140-140 + <18 = 180-140

<18 = 40

8 0

4 years ago

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