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2 years ago

This exact same question has been asked a few times, but I’m finding several different answers. Could someone please clarify.

Mathematics

2 answers:

MA_775_DIABLO [31]2 years ago

8 0

Answer:

The last one (D)

Step-by-step explanation:

Jonathan would have to check how much he's paid off after each of the months which means he would need to subtract that from total amount of money that he's pay for the new computer

Mama L [17]2 years ago

3 0

The correct answer is D) Y= 1,728-96x

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What is the best estimate of the product 3780 times 2

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Answer:

7560

Step-by-step explanation:

I used google calculator

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Aaliyah practices clarinet a total of 11.25 hours in d days. She practices the same amount of time each day. How many hours does

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2 years ago

There is a markup of 40% on the original price of a fan. If the fan was sold for $42, what was its original price

Akimi4 [234]

Answer: 21$ that is the answer

Step-by-step explanation:

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2 years ago

Two piecewise functions are shown below. What is the value of

gizmo_the_mogwai [7]

Step-by-step explanation:

I am not sure what your problem here is.

you understand the inequality signs ?

anyway, to get

6×f(-2) + 3×g(1)

we can calculate every part of the expression separately, and then combine all the results into one final result.

f(-2)

we look at the definition.

into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?

is -2 < -2 ? no.

is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.

so, we have to use

1/3 x³

for x = -2 that is

1/3 × (-2)³ = 1/3 × -8 = -8/3

g(1)

again, we look at the definition.

into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?

is 1 > 2 ? no.

is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.

so we have to use

2×|x - 1| + 3

for x = 1 we get

2×0 + 3 = 3

6×f(-2) = 6 × -8/3 = 2× -8 = -16

3×g(1) = 3× 3 = 9

and so in total we get

6×f(-2) + 3×g(1) = -16 + 9 = -7

3 0

1 year 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

2 years ago