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never [62]
3 years ago
13

What equals 18 to the third power

Mathematics
2 answers:
Orlov [11]3 years ago
8 0
18 to the third power is 5832.
salantis [7]3 years ago
7 0

Answer:

5832

Step-by-step explanation:

18 × 18 = 324

324 × 18 = 5832

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Solve for x PLZ HELP!!!
r-ruslan [8.4K]

21


3 times 5 =15

15+6=21



3 0
4 years ago
Read 2 more answers
When a fair dice is thrown, what is the probability of getting a number greater than 4? (Reduce to simplest form)
Y_Kistochka [10]

Answer:

I searched the q and this is what I found

Step-by-step explanation:

Explanation: Number greater than 4 are 5 and 6 . So required probability is 26=13.

7 0
3 years ago
I need to know if this equation would be true or false​
Sati [7]

Answer:

False

Step-by-step explanation:

85 + (-10) is the same as 85 - 10. 85 - 10 is 75, not 95.

5 0
4 years ago
What is the surface area of the triangular prism shown? A triangular prism. The triangle faces have base 14 meters and height 24
Alekssandra [29.7K]

Answer:

1064 m²

Step-by-step explanation:

The surface area of the prism can be gotten by saying

14 * 14 + 14 * 14 + 14 * 24 =

196 + 196 + 336 =

728 m²

Again, after dealing with the rectangle, we then face the triangle.

Area of triangle is

1/2 * 14 * 24

There are 2 triangles so this makes it

2 * 1/2 * 14 * 24 =

336

The surface area is finally

336 + 728 = 1064 m²

Surface area of the triangular prism has been calculated to be 1064 m²

3 0
3 years ago
A university cafeteria line in the student center is a
nirvana33 [79]

Answer:

Lamda= 4 students/min, µ= 5 students/min  

P= Lamda/µ= 4/5= 0.8

a.) Probability that system is empty= P0= 1-P= 1-0.8= 0.2

b.) Probability of more than 2 students in the system= ∑(n=3 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2)= (.2)*(5-  - .2 - (.8)*.2 – (.2)*.8^2))= 0.848

Probability of more than 3 students in the system= ∑(n=4 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2 – (1-P)*P^3)= 0.768

c.) W(q)= Waiting time in Queue= lamda/µ(µ- lamda)= 4/5(1)= 0.8 minutes

d.) L(q)= lamda*W(q)= 4*.8= 3.2 students

e.) L(System)= lamda/(µ-lamda)= 4 students.

f.) If another server with same efficiency as the 1st one is added, then µ= 6 sec/student= 10 students/min.

P= 4/10= 0.4

Probability that system is empty= P0= 1-.4= 0.6

W(q)= 4/10(10-4)= 0.0667 minutes

L(q)= Lamda*W(q)= 4*.0667=0.2668

L(system)= Lamda/(µ-lamda)= 4/6= .667

Step-by-step explanation:

6 0
4 years ago
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