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2 years ago

If 4 pounds of sugur cost $2, the cost of 1 pound of sugar cost

Mathematics

2 answers:

Bad White [126]2 years ago

8 0

Answer:

50 cent

Step-by-step explanation:

viktelen [127]2 years ago

8 0

Answer:

.50¢

Step-by-step explanation:

50 cents = 1 pound

$1.00 = 2 pound

$1.50 = 3 pounds

$2.00 = 4 pounds

basically adding 50 cents equals to $2

You might be interested in

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

What would you do if when you okay so he said yes would go?

Debora [2.8K]

I'm confused but no? lol

6 0

2 years ago

Michael has a weekly food budget of $62. If he plans to budget the same amount for each of the 7 days of the week, what is the m

garri49 [273]

Answer:

62/7

Step-by-step explanation:

8 0

2 years ago

Jenny’s mother is 5 years older than twice Jenny's age. The sum of their ages is 62 years. This is represented by the equation x

Salsk061 [2.6K]

Answer:

Jenny´s age is 19 and her mother is 43.

Step-by-step explanation:

1. Plug in the first value of the set

41 is different from 62, so Jenny isn´t 12 years old.

2. Plug in the second value of the set

50 is different from 62, so Jenny isn´t 15 years old.

3. Plug in the third value of the set

59 is different from 62, so Jenny isn´t 18 years old.

4. Plug in the fourth value of the set

Therefore Jenny´s 19 years old.

5. Calculate Jenny´s mother age:

As the sum of their ages is 62 ages, we have the following equation

where

x=Jenny´s age

y=Jenny´s mother age

Therefore Jenny's mother age is 43 years old.

pls mark me brainliest

8 0

2 years ago

An equation was created for the line of best fit from actual enrollment data. It was used to predict the dance studio enrollment

stealth61 [152]

Answer: First option is correct.

Step-by-step explanation:

Enrollment month Actual Predicted Residual

January 500 8 4

February 400 15 -1

March 550 15 -1

April 13 12 -1

May 16 17 -1

June 14 15 -1

Since we know that

Residual value = Actual value - Predicted value

Sum of residuals is given by

$4-1-1+1-1-1=1$

since we can see that sum of residual is more than 0.

So, it can't be a good fit .

Hence, No, the equation is not a good fit because the sum of the residuals is a large number.

Therefore, First option is correct.

6 0

2 years ago