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BARSIC [14]
1 year ago
13

A baseball stadium has 45,000 seats. If tickets have been brought for 5/8 of the seats for the next game, how many seats are sti

ll available
Mathematics
1 answer:
Nadusha1986 [10]1 year ago
6 0

Subtraction is a mathematical operation that reflects the removal of things from a collection. The number of seats that are available in the stadium is 16,875.

<h3>What is subtraction?</h3>

Subtraction is a mathematical operation that reflects the removal of things from a collection. The negative symbol represents subtraction.

Given that a baseball stadium has 45,000 seats, out of this 45,000 seats, 5/8 of the seats are bought for the next game.

Therefore, the number of seats that are still available is,

Number of seats available

= Total number of seats - Seats already sold

= 45,000 - (45,000 × 5/8)

= 45,000 - 28,125

= 16,875

Hence, the number of seats that are available in the stadium is 16,875.

Learn more about Subtraction here:

brainly.com/question/1927340

#SPJ1

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Im putting all the questions in one to make it easier
Rufina [12.5K]

I'll focus on problem 19 and problem 22

============================================================

Problem 19

Answers:

  • x = 87
  • x+15 = 102

-------------------------------

Explanation:

The horizontal lines are parallel. Note the arrow markers along the interior portion of the lines. These similar markers tell us how the lines pair up to be parallel to one another.

Since the horizontal lines are parallel, this means the angles in question are congruent. They are corresponding angles.

x+15 = 102

x = 102-15

x = 87

Which leads to

x+15 = 87+15 = 102

============================================================

Problem 22

Answer: y = 27

-------------------------------

Explanation:

We have a pair of alternate exterior angles. They are congruent due to the parallel lines.

Let's solve for y.

7y-55 = 3y+53

7y - 3y = 53+55

4y = 108

y = 108/4

y = 27

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Step-by-step explanation:

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The pointer of the vibration measuring instrument is observed to move between the 0.1 and0.3 marks on the vertical scale, when s
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W = 1000mm/1m = 20,000N/M

FormulaX = FolK/1 - (w/wn)²

0.2 = b/20000/1 - (100/50)²

Given forcing frequency was doubled, W1 = 2 x 100=200rad/s

X1 =b/20,000/1 - (200/20)²

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