If ED = 10 and DB = 3x – 8, what is the length of mc084-2.jpg?

Find the value of x

allsm [11]

Answer:

136

Step-by-step explanation:

x = 1/2 (arc(XY) + arc(WZ) )

Arc (XY ) = 142

Arc(WZ) = 130

x = 1/2 (142 + 130)

x = 1/2 (272)

x = 136

3 0

3 years ago

Which scenario is an example of market saturation? The single coffee shop in town usually has a long line of customers and tends

grin007 [14]

The three shoe shops would be the best example of market saturation because there is too many stores and not enough customers, therefore the market is over saturated with shoe stores.

5 0

3 years ago

Read 2 more answers

What is (x+5)+2x^2=2x+1

andrew-mc [135]

Answer:

Step-by-step explanation:

Area of Square = (x+5)²= x² + 10x + 25

Area of rectangle = 2x * x = 2x²

area of rectangle + 1 = Area of square

2x² + 1 = x² + 10x +25

2x² + 1 - x² - 10x - 25 = 0

x² - 10x - 24 = 0

(x - 12) (x + 2) = 0

x - 12 = 0 ; Ignore x + 2 because length wont come in negative

x = 12

5 0

3 years ago

Read 2 more answers

HELP ME [PLEASE ;) CCCCCCCCC

hjlf

Answer:

its equal 30

Step-by-step explanation:

5 0

3 years ago

The computers of six faculty members in a certain department are to be re- placed. two of the faculty members have selected lapt

FinnZ [79.3K]

Answer:

A) The probability that both selected setups are for laptop computers is 0.067

B)The probability that both selected setups are desktop machines is 0.4

C)The probability that at least one selected setup is for a desktop computer is 0.933

D)The probability that at least on computer of each type is chosen for setup is 0.533

Step-by-step explanation:

Number of laptops = 2

Number of desktops = 4

Total number of outcomes = 15

a) what is the probability that both selected setups are for laptop computers?

Total number of outcomes = 15

So, the probability that both selected setups are for laptop computers = $\frac{1}{15}=0.067$

b)what is the probability that both selected setups are desktop machines?

Number of desktops = 4

Number of desktops to be chosen = 4

We will use combination

No. of ways to select two desktops = $^4C_2=\frac{4!}{2!(4-2)!}=6$

So,the probability that both selected setups are desktop machines= $\frac{6}{15}=0.4$

(c) what is the probability that at least one selected setup is for a desktop computer?

P(at least 1 desktop)=1-P(No desktop)

P(at least 1 desktop)=1-P(both laptops)

P(at least 1 desktop)=1-0.067=0.933

So,the probability that at least one selected setup is for a desktop computer is 0.933

d) what is the probability that at least on computer of each type is chosen for setup?

No. of ways to select one desktop $=^4C_1=\frac{4!}{1!(4-1)!}=4$

No. of ways to select one laptop = $^2C_1=\frac{2!}{1!(2-1)!}=2$

So, No. of ways to select one laptop and one desktop= $4 \times 2 = 8$

So,the probability that at least on computer of each type is chosen for setup= $\frac{8}{15}=0.533$

6 0

3 years ago