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3 years ago

What is the slope ? Please help

Mathematics

2 answers:

S_A_V [24]3 years ago

4 0

Answer:

y = -2/3x + 3 pic below to show proof i hope this helps! :)

Step-by-step explanation:

looking at the photo this slope is negative

we can see 2 given points (0,3) and (3,1)

y = +/- ?/?x +/- ?

from the point (0,3) we move down 2

y = -2/? x + ?

from the point that was moved down 2, we move over 3 to get to (3,1) y = -2/3x + ?

the line crosses the 3 on the y-axis

y = -2/3x + 3

lisov135 [29]3 years ago

3 0

Answer:

-2/3

Step-by-step explanation:

Okay, this is a fairly simple question.

First, you want to pick two pairs of coordinates (X,Y).

In this case, I chose (-3, 5) and (3, 1)

Now, you want to use the formula to find slope : y2-y1/x2-x1

1-5/3-(-3)

-4 divided by 6

=-2/3

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1 year ago

With a height of 68 in, Nelson was the shortest president of a particular club in the past century. The club presidents of the

Ivahew [28]

Answer:

a. The positive difference between Nelson's height and the population mean is: $\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).

c. Nelson's z-score: $\\ z = -1.1739 \approx -1.174$ (Nelson's height is below the population's mean 1.174 standard deviations units).

d. Nelson's height is usual since $\\ -2 < -1.174 < 2$ .

Step-by-step explanation:

The key concept to answer this question is the z-score. A z-score "tells us" the distance from the population's mean of a raw score in standard deviation units. A positive value for a z-score indicates that the raw score is above the population mean, whereas a negative value tells us that the raw score is below the population mean. The formula to obtain this z-score is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where

$\\ z$ is the z-score.

$\\ \mu$ is the population mean.

$\\ \sigma$ is the population standard deviation.

From the question, we have that:

Nelson's height is 68 in. In this case, the raw score is 68 in $\\ x = 68$ in.
$\\ \mu = 70.7$ in.
$\\ \sigma = 2.3$ in.

With all this information, we are ready to answer the next questions:

a. What is the positive difference between Nelson's height and the mean?

The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):

$\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in$ .

That is, the positive difference is 2.7 in.

b. How many standard deviations is that [the difference found in part (a)]?

To find how many standard deviations is that, we need to divide that difference by the population standard deviation. That is:

$\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174$

In words, the difference found in part (a) is 1.174 standard deviations from the mean. Notice that we are not taking into account here if the raw score, x, is below or above the mean.

c. Convert Nelson's height to a z score.

Using formula [1], we have

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}$

$\\ z = \frac{-2.7\;in}{2.3\;in}$

$\\ z = -1.1739 \approx -1.174$

This z-score "tells us" that Nelson's height is 1.174 standard deviations below the population mean (notice the negative symbol in the above result), i.e., Nelson's height is below the mean for heights in the club presidents of the past century 1.174 standard deviations units.

d. If we consider "usual" heights to be those that convert to z scores between minus2 and 2, is Nelson's height usual or unusual?

Carefully looking at Nelson's height, we notice that it is between those z-scores, because:

$\\ -2 < z_{Nelson} < 2$

$\\ -2 < -1.174 < 2$

Then, Nelson's height is usual according to that statement.

7 0

2 years ago

SOMEONE HELP MEEEEEE 75 POINTS TO THE PERSON THAT HELPS

Tresset [83]

Answer:

Part 1) $9x-7y=-25$

Part 2) $2x-y=2$

Part 3) $x+8y=22$

Part 4) $x+8y=35$

Part 5) $3x-4y=2$

Part 6) $10x+6y=39$

Part 7) $x-5y=-6$

Part 8)

case A) The equation of the diagonal AC is $x+y=0$

case B) The equation of the diagonal BD is $x-y=0$

Step-by-step explanation:

Part 1)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M=(\frac{2-6}{2},\frac{-3+5}{2})$

$M=(-2,1)$

step 2

The equation of the line into point slope form is equal to

$y-1=\frac{9}{7}(x+2)\\ \\y=\frac{9}{7}x+\frac{18}{7}+1\\ \\y=\frac{9}{7}x+\frac{25}{7}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=\frac{9}{7}x+\frac{25}{7}$

Multiply by 7 both sides

$7y=9x+25$

$9x-7y=-25$

Part 2)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M=(\frac{1+5}{2},\frac{0-2}{2})$

$M=(3,-1)$

step 2

Find the slope

The slope between two points is equal to

$m=\frac{-2-0}{5-1}=-\frac{1}{2}$

step 3

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

Find the slope of the line perpendicular to the segment joining the given points

$m1=-\frac{1}{2}$

$m1*m2=-1$

therefore

$m2=2$

step 4

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=2$ and point $(1,0)$

$y-0=2(x-1)\\ \\y=2x-2$

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=2x-2$

$2x-y=2$

Part 3)

In this problem AB and BC are the legs of the right triangle (plot the figure)

step 1

Find the midpoint AB

$M1=(\frac{-5+1}{2},\frac{5+1}{2})$

$M1=(-2,3)$

step 2

Find the midpoint BC

$M2=(\frac{1+3}{2},\frac{1+4}{2})$

$M2=(2,2.5)$

step 3

Find the slope M1M2

The slope between two points is equal to

$m=\frac{2.5-3}{2+2}=-\frac{1}{8}$

step 4

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{8}$ and point $(-2,3)$

$y-3=-\frac{1}{8}(x+2)\\ \\y=-\frac{1}{8}x-\frac{1}{4}+3\\ \\y=-\frac{1}{8}x+\frac{11}{4}$

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=-\frac{1}{8}x+\frac{11}{4}$

Multiply by 8 both sides

$8y=-x+22$

$x+8y=22$

Part 4)

In this problem the hypotenuse is AC (plot the figure)

step 1

Find the slope AC

The slope between two points is equal to

$m=\frac{4-5}{3+5}=-\frac{1}{8}$

step 2

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{8}$ and point $(3,4)$

$y-4=-\frac{1}{8}(x-3)$

$y=-\frac{1}{8}x+\frac{3}{8}+4$

$y=-\frac{1}{8}x+\frac{35}{8}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=-\frac{1}{8}x+\frac{35}{8}$

Multiply by 8 both sides

$8y=-x+35$

$x+8y=35$

Part 5)

The longer diagonal is the segment BD (plot the figure)

step 1

Find the slope BD

The slope between two points is equal to

$m=\frac{4+2}{6+2}=\frac{3}{4}$

step 2

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$ and point $(-2,-2)$

$y+2=\frac{3}{4}(x+2)$

$y=\frac{3}{4}x+\frac{6}{4}-2$

$y=\frac{3}{4}x-\frac{2}{4}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=\frac{3}{4}x-\frac{2}{4}$

Multiply by 4 both sides

$4y=3x-2$

$3x-4y=2$

Note The complete answers in the attached file

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3 0

3 years ago