What is the base of x^4?<br><br> a. 4<br> b. x<br> c. 1

What is the distance between points (8,3) and (53,31) on a coordinate plane?

Norma-Jean [14]

Answer:

53 units

Step-by-step explanation:

Let $(x_1,y_1)\rightarrow(8,3)$ and $(x_2,y_2)\rightarrow(53,31)$ to use the distance formula:

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\d=\sqrt{(31-3)^2+(53-8)^2}\\\\d=\sqrt{(28)^2+(45)^2}\\\\d=\sqrt{784+2025}\\\\d=\sqrt{2809}\\\\d=53$

Thus, the distance between the two points is 53 units

7 0

2 years ago

Determine whether 2/15 is closest to 0,1/2 or 1

stira [4]

Since 15ths are smaller pieces 2 of 15 would'nt even really be close to 1/2 half of 15 is 7.5 so not even close. or 1 so the answer would be that it is closer to 0. Hope it helps!!

7 0

4 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

What is this thing with the 3 red blades???

zubka84 [21]

oumm I already saw this in our mall also in the tv... NAME OF THIS TOY WAS "BEYBLADE"

Step-by-step explanation:

mm! this is a toy for kids...

for making them not bored..

3 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST! HELP!

ki77a [65]

Answer:

(ii) is correct, please proceed.

Step-by-step explanation:

5 0

3 years ago

What is the base of x^4? a. 4 b. x c. 1