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Dvinal [7]
3 years ago
11

A performer expects to sell 5000 tickets for an upcoming concert. They plan to make a total of 311000 in sales from these ticket

s. Assume that all tickets have the same price. How much money will they make if they sell 7000 tickets?
Mathematics
1 answer:
Marina86 [1]3 years ago
6 0
435400 i think,as one ticket costs 62.2 in this case which means if we do 7000 multiplied by 62.2 ,we get435400
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Mikhael bought refreshments. Sandwiches were $3 each and bottles of water were $0.75 each. He spent $318.75 on a total of 200 it
Anestetic [448]
X= sandwiches
y= bottles of water

QUANTITY EQUATION
x+y= 200

COST EQUATION
$3x + $0.75y= $318.75

STEP 1
solve for one variable in quantity equation and substitute in cost equation

x+y= 200
subtract y from both sides
x= 200-y

STEP 2
substitute 200-y in for x

$3x + $0.75y= $318.75
3(200-y) + 0.75y= 318.75
600 - 3y + 0.75y= 318.75
combine like terms
600 - 2.25y= 318.75
-2.25y= -281.25
divide both sides by -2.25
y= 125 number of water bottles

STEP 3
substitute y=125 in either equation

x+y= 200
x + 125= 200
x= 75 sandwiches

ANSWER: There were 75 sandwiches and 125 bottles of water.

Hope this helps! :)
6 0
4 years ago
OMG PLEASE HELP MEEEEEEEEEEEEEEE
katrin [286]
1.2+2.6x1.4-3.9
1.2+3.64-3.9
B) 0.94
3 0
3 years ago
Read 2 more answers
Two chess players, A and B, are going to play 7 games. Each game has three possibleoutcomes: a win for A (which is a loss for B)
Lina20 [59]

Answer:

A) the possible outcomes for individual game is 210 games.

B) The possible outcome for this is 357 games.

C) Then the Possible Games are 267 games.

Step-by-step explanation:

A) Total number of individual games are 7 in which A ends up with 3 wins which give 4 remaining games. then there are two draw games from them. and in the end the remaining games are losses so,we useformula for combination:

                       (7C3)*(4C2)=210 \ games

B) Now Player A has 4 point that gives us 5 possibilities until we reach 7 games. Similarly B have 3 points which is same as player A gives us 5 possibilities until we reach 7 games. thus those two cases for player A and Player B can be stated as:

           (7C3)+(7C3)*(4C2)+(7C1)*(6C6)+(7C2)*(5C4)=357\  games

C) Lets say that player A ahs 4 points and player B has 3 points and all seven games have been played so.

c1) If player A has 4 wins and 3 losses then last win have to be in 7th match thus:the answer is (6C3).

c2) If player A has 3 win, 2 draws and 3 losses thus it means that final match cannot be a loss thus the answer is (6C2)*(5C3).

c3) Now lastly if player A has 1 win and 5 draws we can arrange them arbitrarily thus the answer here is (7C1).

c4) now If player A has 2 wins, 4 draws and 2 losses thus answer is (6C1)*(6C2)

Sum of all the cases is

               (6C3)+(6C2)*(5C3)+(7C1)+(6C1)*(6C2)=267\ games

4 0
3 years ago
Find the standard deviation, _, for the binomial distribution which has the stated values of n and p. n=38; p=2/5
NARA [144]

Answer:

Solution: given that

n = 38

p = 2/5 = 0.4 q = 1-p = 1-0.4 = 0.6

standard deviation = sqrt(n*p*q)

sqrt(38*0.4*0.6) = 3.0199

5 0
3 years ago
PLEASE HELP!!! 15 POINTS
Alina [70]

Answer:

A=150(1+2)^4

Step-by-step explanation:

Look up the compound interest formula, this is how you solve the problem.

A=P(1+r/n)^(nt)

A=150(1+(300%-100%)/1)^(1)4

A=150(1+2)^4

A=12150

If you have any questions about the math please feel free to ask. Have a nice day.

3 0
3 years ago
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