When seven times a number is decreased by 3, the result is 39. What is the number?

Archy [21]

From what i see, i guess the answer may be B. it looks like it, but i dont really know.

6 0

3 years ago

Which of the following could be the number of real roots, including any repeated roots, of a quadratic polynomial with real coef

scoray [572]

I think the only answer here is 0
Idk tho

5 0

3 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

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

What is the slope of the line in the graph?

Bingel [31]

Answer:

The slope equals the rise divided by the run:

7 0

3 years ago

Can anybody help me with 1 and 2

Anit [1.1K]

1) $x \geq 96$
2) $x\ \textless \ 38$

Here, the variable $x$ can assume any value, and usually an inequality presented like this has $x$ as a stand-in for all real numbers. If we really wanted to make that fact clear, we could write the two inequalities like this:

$x \geq 96, x\in\mathbb{R}\\ x\ \textless \ 38, x\in\mathbb{R}$

Formally, you'd read $x\in\mathbb{R}$ as " $x$ is an element of the set of real numbers," which is a fancy way of saying that $x$ can be any real number.

8 0

3 years ago