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3 years ago

In Prof. Johnson's class, the 10 students had the following scores on the last midterm.

Mathematics

1 answer:

timofeeve [1]3 years ago

7 0

Answer: 188 is the outlier

Step-by-step explanation:

Find median of the whole data set. The median is 83.5.

Then find Q1 and Q3. Q1 is 78, Q2 is 96.

Then find the IQR. The IQR is 18.

Find the inner fences of the data set. The inner fences are 51 and 123. Anything that is outside of 51 or 123 is an outlier so 188 is an outlier.

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The temperature changed from 7°F at 6 p.M to -5°F at midnight. What was the difference between the high and low temperatures? Wh

IgorC [24]

Answer:

12 °F

2°F per hour

Step-by-step explanation:

The difference can be found by substracting:

7°F-( -5 °F)

=12 °F

To find the average change in temperature per hour, divide by how many hours between 6 to midnight. Between 6 to midnight is 6 hours.

12°F / 6 hours

= 2°F/ hour

7 0

3 years ago

= 9x2 +9x -18,/<br> 3x +6

Ede4ka [16]

Answer:24+3x

Step-by-step explanation:

maybee i just used photomath

3 0

3 years ago

Circle the largest decimal in each of the following groups.

ExtremeBDS [4]

A. 0.370

B. 0.44

C. 0.1

D. 2.75

7 0

3 years ago

Read 2 more answers

Y = 3x - 12x +9<br> In standard form?

xxMikexx [17]

Answer:

y=-9x+9

Step-by-step explanation:

There are only two like terms in the equation which I have underlined.

y= <u>3x - 12x</u> +9

To simplify it, you combine the like terms and get your equation in standard form.

5 0

2 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