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3 years ago

Write the line (-5,-6) with slope 2/5 in point-slope form

Mathematics

1 answer:

stich3 [128]3 years ago

5 0

Answer:

$y+6=\frac{2}{5} (x+5).$

Step-by-step explanation:

1) the common form of the point-slope form is y-y₁=s(x-x₁), where s - the slope, (x₁;y₁) - the point belongs to the required line;

2) if s=2.5; x₁= -5; y₁= -6, then the required equation is:

y+6=2/5(x+5).

You might be interested in

Select the condition for which it is NOT possible to construct a triangle.

maw [93]

Answer:

A triangle with side lengths 4 cm, 5 cm, and 15 cm

Step-by-step explanation:

In order to form a triangle with given side lengths, sum of any two sides must always be greater than the third side.

$4+5= 9\ngtr 15\\$

6 0

3 years ago

Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. , The mean, ,

klemol [59]

Complete Question

Find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p. , The mean, , is nothing. (Round to the nearest tenth as needed.)

p = 0.6 n = 18

Answer:

The mean $\mu = 10.5$

The standard deviation $\sigma = 2.08$

The variance $var = 4.32$

Step-by-step explanation:

From the question we are told that

The probability of success is $p = 0.6$

The sample size is $n = 18$

Generally given that the distribution is binomial, then the probability of failure is mathematically represented as

$q = 1- p$

substituting values

$q = 1- 0.6$

$q =0.4$

Generally the mean is mathematically evaluated as

$\mu = np$

substituting values

$\mu = 18 * 0.6$

$\mu = 10.5$

The standard deviation is evaluated as

$\sigma = \sqrt{npq}$

substituting values

$\sigma = \sqrt{18 * 0.6 * 0.4}$

$\sigma = 2.08$

The variance is evaluated as

$var = \sigma^2$

substituting value

$var = 2.08^2$

$var = 4.32$

4 0

4 years ago

What is the radius of a semi circle when the area is 20cm

vazorg [7]

hello :
the area of the circle is : 40 cm²
40 = <span>πr²
r² = 40/(3.14 )
r² = 12.73 cm</span>

6 0

3 years ago

4:3 = 60:45 true or false

hjlf

True multiply both sides by 15

3 0

4 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