Find the distance between the parallel lines y=x+8 and y=x+2

Based on what you know what is the measure of m

Bond [772]

Answer:

59 degrees, 44 plus 77 is 121, subtract from 180

4 0

3 years ago

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

2)

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

How many right angles are there in the diagram?<br> There are<br> right angles.

TiliK225 [7]

I think is 4 follow me and la

4 0

3 years ago

-4x -7 - 3x + 4 = 25

zimovet [89]

Answer:

x=-4

Step-by-step explanation:

-4x -7 - 3x + 4 = 25

-7x=25-4+7

x=-28/7

x=-4

7 0

3 years ago

Read 2 more answers

The power P of a jet of water is jointly proportional to the cross sectional area A and to the cube ofthe velocity V. (a) Write

anastassius [24]

Answer:

a) $P=kAV^3$

b) 54

Step-by-step explanation:

a)

If one quantity, A, is jointly proportional to other 2 quantities (B & C), we can write proportionality equation as:

$A=k*B*C$

Where

k is the proportionality constant

Now, for this problem, P is jointly proportional to A and CUBE OF V, so we can write:

$P=kAV^3$

b)

Now velocity, V, is triples, which means it becomes 3V

Also, cross sectional area, A, is doubled, which means it becomes 2A

Let's put these into formula and see the difference:

$P=kAV^3\\P=k(2A)(3V)^3\\P=k(2A)(27V^3)\\P=(2*27)kAV^3\\P=54kAV^3$

So, power is multiplied by a factor of 54. Power becomes 54 times if Area is doubled and velocity is tripled ( as seen ).

8 0

2 years ago