All categories

2 years ago

A family has a monthly income of $3912. Their monthly expenditures are as shown in the graph below.

Mathematics

1 answer:

stealth61 [152]2 years ago

5 0

Answer: The save $1134.48

Step-by-step explanation:

3912 x 0.29 = 1134.48

You might be interested in

9 ( x + 3 ) - (2x + 5) = *

Ivan

Answer:

////////////////////////////////////

=7x + 22

7 0

4 years ago

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

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.

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.

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.

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

Solve the following equation for x to find the total number of people who became members of a social networking site for a certa

zavuch27 [327]

Answer:

360 people

Step-by-step explanation:

x = 0.8x + 72 Subtract 0.8x on both sides

0.2x = 72 Divide 0.2 on both sides

x = 360

5 0

4 years ago

Read 2 more answers

The measures of the angles of a triangle are in the ratio 2:3:4. In degrees, the measure of the largest angle of the triangle is

notsponge [240]

Since the three angles of a triangle add to give 180°

The the size of the largest angle 4 = 180° × $\frac{4}{2+3+4}$
= 180° × $\frac{4}{9}$

∴ the size of the largest angle = 80°

8 0

3 years ago

(x^2 - x^(1/2))/(1-x^(1/2))

Levart [38]

$\frac { \left( { x }^{ 2 }-{ x }^{ \frac { 1 }{ 2 } } \right) }{ \left( 1-{ x }^{ \frac { 1 }{ 2 } } \right) }$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot 1$

$\\ \\ =\frac { \left( { x }^{ 2 }-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot \frac { \left( 1+\sqrt { x } \right) }{ \left( 1+\sqrt { x } \right) }$

$\\ \\ =\frac { { x }^{ 2 }+{ x }^{ 2 }\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }$

$\\ \\ =\frac { -\sqrt { x } \left( 1-{ x }^{ 2 } \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\\ \\ =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\} }{ \left( 1-x \right) }$

$\\ \\ =-\sqrt { x } \left( 1+x \right) -x\\ \\ =-{ x }^{ \frac { 1 }{ 2 } }\left( 1+{ x }^{ \frac { 2 }{ 2 } } \right) -x$

$\\ \\ =-{ x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }-x\\ \\ =-\sqrt { x } -\sqrt { { x }^{ 3 } } -x$

3 0

3 years ago

Read 2 more answers

A family has a monthly income of ​$3912. Their monthly expenditures are as shown in the graph below.

A family has a monthly income of $3912. Their monthly expenditures are as shown in the graph below.