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3 years ago

Find the midpoint of the line segment that has endpoints at (10,8) (-3,-10)

Mathematics

1 answer:

dem82 [27]3 years ago

7 0

Hey there! :)

Answer:

(3.5, -1).

Step-by-step explanation:

Use the midpoint formula to solve this problem:

$(x_m, y_m) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Plug in the coordinates given:

$(\frac{10-3}{2}, \frac{8-10}{2})$

Simplify:

$(\frac{7}{2}, \frac{-2}{2})$

Therefore, the coordinates of the mid-point are:

(3.5, -1).

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Question 9 (1 point)

Romashka [77]

Answer:

Step-by-step explanation:

League A League B

151.12 163.25

148 157

26.83 24.93

29 136

136 145

167 178

207 256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(82.5-1.5\times 76.56,159.06+1.5\times 76.56)$

(-32.34,273.9)

So, There is no outlier

Maximum = 207

League B in ascending order :

24.93,136,145,157,163.25,178,256

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

$Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5$

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(140.5-1.5\times 30.125,170.625+1.5\times 30.125)$

(95.3125,215.8125)

24.93 and 256 are outliers

Maximum = 256

5 0

3 years ago

16. Tell whether the lines for each pair of equations are parallel, perpendicular, or neither.

il63 [147K]

Answer:

Given equation are:

$y = -\frac{1}{4}x+8$ ......[1]

$-2x+8y = 4$ .....[2]

The two lines are parallel lines then their slopes will be equal.

When two lines are perpendicular then, the slope of lines are the negative reciprocals of each other.

Now, Equation of a line is in the form of y =mx+b where m is the slope of the line.

Slope $(m_1)$ of equation of line in [1] is;

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

then;

$m_1= -\frac{1}{4}$

Slope $(m_2)$ of equation of line in [2];

$-2x+8y = 4$

Add both sides 2x we get;

-2x + 8y + 2x = 2x + 4

Simplify:

8y = 2x +4

Divide both sides by 8 we get;

$y = \frac{1}{4} x + \frac{1}{2}$

then;

$m_2 = \frac{1}{4}$

Therefore, the given two lines are neither parallel nor perpendicular.

4 0

3 years ago

If the radius of a circle is 20 cm, what is the circumference? (Use π = 3.14)

PIT_PIT [208]

Okay so the circumference of a circle is pi×the diameter or 2×radius× pi. In this case we are given the radius so
C=2×20×3.14
C=40×3.14
C=125.60
The circumference is 125.60cm

4 0

3 years ago

Will give brainliest, *15 points*

olganol [36]

2.4

240u^2 / 100u^2 = 2.4

5 0

3 years ago

I need help please i am giving 100 points and brainliest

liubo4ka [24]

Answer:

-3

Step-by-step explanation:

-7x + 4x = 9

-3x = 9

x = 9/(-3)

<h2>MARK ME AS BRAINLIST </h2>

8 0

2 years ago

Read 2 more answers