Is this equation an identity?<br> 10.95 + 14.5n – 19.73 = 14.5n

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla

svp [43]

Here is the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29: 117 122 129 118 131 123

Men aged 60-69: 130 153 141 125 164 139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

$coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = $\dfrac{\sum \limits^{n}_{i-1}x_i}{n}$

Mean = $\frac{117+122+129+118+131+123}{6}$

Mean = $\dfrac{740}{6}$

Mean = 123.33

Standard deviation $= \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }$

Standard deviation = $\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }$

Standard deviation = $\sqrt{\dfrac{161.3334}{5}}$

Standard deviation = $\sqrt{32.2667}$

Standard deviation = 5.68

The $coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

$coefficient \ of \ variation = \dfrac{5.68}{123.33}*100$

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = $\dfrac{\sum \limits^{n}_{i-1}x_i}{n}$

Mean = $\frac{ 130 + 153 + 141 + 125 + 164 + 139}{6}$

Mean = $\dfrac{852}{6}$

Mean = 142

Standard deviation $= \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }$

Standard deviation = $\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }$

Standard deviation = $\sqrt{\dfrac{1048}{5}}$

Standard deviation = $\sqrt{209.6}$

Standard deviation = 14.48

The $coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

$coefficient \ of \ variation = \dfrac{14.48}{142}*100$

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0

3 years ago

Line BD is tangent to circle O at C, Arch AEC = 299, and ACE = 93. Find Angle DCE.

VladimirAG [237]

In triangle ACE,
we know C=93,E can be calulated by using arch angle AEC...what ever that is....,using this we get A=180-(E+93)

So, by alternate segment theorem, DCE= A.

thats all i can say.

7 0

3 years ago

Ginger buys lunch at school every day. She always gets

shepuryov [24]

Answer:

The estimated probability that Ginger will eat a a pizza everyday of the week is;

D. 8/10 = 80%

Step-by-step explanation:

The given parameters are;

The frequency with which Ginger buys launch = Everyday

The percentage of the time the cafeteria has pizza out = 80%

The outcome of 0 and 1 = No pizza available

The outcome of 2, 3, 4, 5, 6, 7, 8, and 9 = Pizza available

Therefore, we have the;

Group number ${}$ Percentage of time pizza available

1 ${}$ 80%

2 ${}$ 80%

3 ${}$ 80%

4 ${}$ 80%

5 ${}$ 40%

6 ${}$ 100%

7 ${}$ 80%

8 ${}$ 100%

9 ${}$ 80%

10 ${}$ 80%

Therefore, the sum of the percentages outcome the days Ginger eats pizza = 0.8 + 0.8 + 0.8 + 0.8 + 0.4 + 1 + 0.8 + 1 + 0.8 + 0.8 = 8

The number of runs of simulation = 10 runs

The estimated probability that Ginger will eat a a pizza everyday of the week = (The sum of the percentages outcome the days Ginger eats pizza)/(The number of runs of simulation)

∴ The estimated probability that Ginger will eat a a pizza everyday of the week = 8/10

6 0

3 years ago

Read 2 more answers

For the concentric circles shown, the radius of the inner circle is 3 centimeters and the distance from the

Tema [17]

Answer:

d

Step-by-step explanation:

Area of the inner circle = 3^2 π = 9 π

total area = 10^2 π = 100π

area of the shaded region

100π - 9 π = 91 π

5 0

3 years ago

Nora can do 70 push ups in 2 minutes. Luke can do 105 push ups in 3 minutes. Do these rates form a proportion? explain

poizon [28]

Yes they do. In order to do this you must make the fraction 70/2 and 105/3. If you divided each one of those you get 35 for both which means it's a proportion.

4 0

3 years ago

Is this equation an identity? 10.95 + 14.5n – 19.73 = 14.5n – 8.78