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Firdavs [7]
3 years ago
14

a popular movie debuted in 2940 theaters and earned an average of $12,490 per cinema the weekend of the debut. Find the amount t

hat the movie earned in its first weekend.
Mathematics
1 answer:
Yuri [45]3 years ago
8 0
Hello! So to find the amount the movie earned that first weekend, all you have to do is multiply the amount of theaters by the amount made per theater. In this case, there are 2,940 theaters and made an average of $12,490. 2,940 * 12,490 is 36,720,600. There. The movie made $36,720,600 on it's opening weekend.
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Find the cube roots of 125(cos 288° + i sin 288°)
kow [346]
Let r(cos O + i sin O)  be a cube root of 125(cos 288 + i sin 288)
then
r^3(cos O + i sin O)^3  =  125(cos 288 + i sin 28)

so r^3 = 125  and  cos 3O + i sin 3O  =  cos 288 + i sin 288

so r  = 5  and 3O = 288 + 360p and O = 96 +  120p

so one cube root is   5 (cos 96 + i sin 96)

Im a little rusty at this stuff Its been a long time.

Im not sure of the other 2 roots

sorry cant help you any more


3 0
3 years ago
Read 2 more answers
A campus deli serves 101 customers over its busy lunch period from 11:30 am to 1:30 pm. A quick count of the number of customers
iogann1982 [59]

Answer:

21.38 minutes

Step-by-step explanation:

the total number served = 101

teh number of hours that is spent in serving from 11.30 to 1.30 = 2 hours

Lq = average number of those that are in line = 18 customers

from here we are to find the flow rate

= \frac{total served}{service hour}

= 101/2 = 50.5

the average time in waiting =

Lq/flow rate

= 18/50.5

= 0.3564 hours

convert to minutes = 0.3564 * 60 minutes

= 21.38 minutes

4 0
3 years ago
3(x +2)-9x+5<br> What is the answer
Alika [10]

Answer:

-6x+11

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
I AM GIVING 45 POINTS TO WHOEVER GETS THIS RIGHT... plz answer corectly and try... i was working on this for sooo long. Plz try
Alexxx [7]
When you have 3 choices for each of 6 spins, the number of possible "words" is
  3^6 = 729

The number of permutations of 6 things that are 3 groups of 2 is
  6!/(2!×2!×2!) = 720/8 = 90

A) The probability of a word containing two of each of the letters is 90/729 = 10/81


The number of permutations of 6 things from two groups of different sizes is
  (2 and 4) : 6!/(2!×4!) = 15
  (3 and 3) : 6!/(3!×3!) = 20
  (4 and 2) : 15
  (5 and 1) : 6
  (6 and 0) : 1

B) The number of ways there can be at least 2 "a"s and no "b"s is
  15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.


_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
6 0
3 years ago
Giving out extra points please help
andreyandreev [35.5K]

Answer:

i dont want extra one i only want the though one

5 0
3 years ago
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