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3 years ago

PLEASE HELPPPPPPLPPPPPPPP

Mathematics

1 answer:

solong [7]3 years ago

3 0

<h3>Answer: 36%</h3>

============================================

Explanation:

There are 20 blue cards out of 10+20+25 = 55 cards total.

Divide the 20 over 55 to get

20/55 = 0.363636...

where the "36"s go on forever

So 20/55 = 0.3636 approximately

Move the decimal point over 2 spots to go from 0.3637 to 36.37%

Then round to the nearest percent: 36.37% ---> 36%

The probability of getting a blue card is approximately 36%

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Mei Su had 80 coins. She gave most of them to her friends in such a way that each of her friends got at least one coin and no tw

Mama L [17]

Answer: 12 friends.

Step-by-step explanation:

the data we have is:

Mei Su had 80 coins.

She gave the coins to her friends, in such a way that every friend got a different number of coins, then we have that:

The maximum number of friends that could have coins is when:

friend 1 got 1 coin

friend 2 got 2 coins

friend 3 got 3 coins

friend N got N coins

in such a way that:

(1 + 2 + 3 + ... + N) ≥ 79

I use 79 because "she gave most of the coins", not all.

We want to find the maximum possible N.

Then let's calculate:

1 + 2 + 3 + 4 + 5 = 15

15 + 6 + 7 + 8 + 9 + 10 = 55

now we are close, lets add by one number:

55 + 11 = 66

66 + 12 = 78

now, we can not add more because we will have a number larger than 80.

Then we have N = 12

This means that the maximum number of friends is 12.

5 0

3 years ago

Need help finding the parallel and perpendicular slope

Rudik [331]

Answer:

Step-by-step explanation:

(x₁, y₁) = (19 , -4) & (x₂ ,y₂) = (17, -20)

$Slope =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$= \frac{-20-[-4]}{17-19}\\\\= \frac{-20+4}{17-19}\\\\= \frac{-16}{-2}\\\\= 8$

m = 8

Parallel lines have same slope.

Parallel slope = 8

Slope of perpendicular line = $\frac{-1}{m}$

Perpendicular slope = $\frac{-1}{8}$

8 0

3 years ago

Which of the following are NOT congruent?

algol13

the adjacent angles in a parallelogram

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

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

Please Help With this question!<br> There are 3 pictures!

Likurg_2 [28]

Corresponding Angles postulate is ur answer mate

8 0

2 years ago