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scoundrel [369]
2 years ago
13

A total of 243 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

ickets sold was two
times the number of adult tickets sold. How many adult tickets were sold?
Mathematics
2 answers:
alina1380 [7]2 years ago
8 0

Answer:

81 adult tickets

Step-by-step explanation:

Step 1:

Set up a system of equations

Since the adult tickets and student tickets add to 243, we can do

a+s=243

and the student tickets sold were 2x the adult tickets, we can do

s=2a

We can replace s in the first equation adding to 243 with 2a

a+s(2a)=243\\3a=243\\

Step 2:

Solve for a. We can divide both sides by 3.

a=81

Therefore, there were 81 adult tickets sold.

Step 3:

We can check our work! Since we learned earlier the student tickets are 2x adult tickets, we can multiply 81 by 2 and add that to 81 to see if it equals 243!

s=2a\\s=2(81)\\s=162

And...

162+81=243

So, 81 must be the answer!

Serggg [28]2 years ago
5 0

Answer:

ok so I did this before

Step-by-step explanation:

just multiply 243 × 2 = ?

I got it right so you will probably get it right :)

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The side length is 5 units, and the area is 25 square units

<h3>What are areas?</h3>

The area of a shape is the amount of space it occupies

<h3>What is the area of a square</h3>

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From the graph, the coordinates of the square can be assumed to be:

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So, we have:

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sertanlavr [38]

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zaharov [31]

Answer:

1. a. 25(1) + 10(2) + 3(3) + 10(4) = 94/140 = .671

2. b. Mark’s average < Jay’s average

Step-by-step explanation:

1. What is Mark’s slugging average?

The slugging average gives more weight to the multi-base hits (compared to single-base hits) in opposition with the batting average.

So, a single hit is worth 1 point, a double is worth 2 points, a triple 3 points and a homerun is worth 4 points.  It's a weighted average calculation.

Mark had 25 singles, 10 doubles, 3 triples and 10 homeruns during 140 presences 140 at bat.

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2. How does Mark's average compares to Jay's?

Let's first calculate Jay's slugging average then we'll be able to decide.

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10 (1) + 6 (2) + 5 (3) + 14 (4) = 10+12+15+56 = 93 / 90 = 1.033

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b. Mark’s average < Jay’s average

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