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sasho [114]
3 years ago
15

A school group wants to rent part of a bowling alley to have a party. The bowling alley costs $500 to rent, plus an additional c

harge of $5 per student to bowl. The group doesn’t want any student to pay more than $15 total to attend this party.
What is inequality that could represent this situation?
How many students would need to attend so each student would pay at most $15?
Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

If there are N students, the total cost will be:

C(N) = $500 + $5*N.

And we want the amount that each student pays to be equal or less than $15.

then:

C(N)/N ≤ $15.

Then the inequality that represents this situation is:

($500 + $5*N)/N ≤ $15.

Now, let's solve this:

($500 + $5*N) ≤ $15*N

$500 ≤ $15*N - $5*N

$500 ≤ $10*N

$500/$10 ≤ N

50 ≤ N

So the minimum number of students needed is 50.

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Here's an explanation! :)

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3 years ago
Read 2 more answers
Answer ,how to do it.
guapka [62]

Answer:

<u>The system has two solutions:</u>

<u>x₁ = 5  ⇒ y₁ = -10</u>

<u>x₂ = -2 ⇒ y₂ = 11</u>

Step-by-step explanation:

Let's solve the system of equations, this way:

y = -3x + 5

y = x ² - 6x - 5

Replacing y in the 2nd equation:

y = x ² - 6x - 5

-3x + 5 = x ² - 6x - 5

x ² - 3x - 10 = 0

Solving for x, using the quadratic formula:

(3 +/- √(9 -4 * 1 * -10))/2 * 1

(3 +/- √9 + 40)/2

(3 +/- √49)/2

(3 +/- 7)/2

x₁ = 10/2 = 5

x₂ = -4/2 = -2

x₁ = 5  ⇒ y₁ = -10

x₂ = -2 ⇒ y₂ = 11

<u>As we can see the system has two different solutions</u>

8 0
4 years ago
The sale price of a table is $472 it was 20% off what was the original price?
ira [324]
To solve, set an equation:

472=0.8x

Divide both sides by 0.8

472/0.8=0.8x/0.8

x=590

Answer: The original price was $590
5 0
3 years ago
PLEASE HELP ASAP!! CORRECT ANSWERS ONLY PLEASE!!!
rosijanka [135]

Answer:

c

Step-by-step explanation:

kzjzjxjxjxjxjxjdjd

6 0
3 years ago
In the diagram above​
Reil [10]

Answer:

<h2>40°</h2>

Step-by-step explanation:

We have parallel lines.

Angles 1 and 2 are alternate angles.

With parallel lines, alternate angles are congruent.

Therefore

m\angle1=m\angle2

m\angle1=2x+20,\ m\angle2=3x+10\\\\2x+20=3x+10\qquad\text{subtract 20 from both sides}\\\\2x+20-20=3x+10-20\\\\2x=3x-10\qquad\text{subtract}\ 3x\ \text{from both sides}\\\\2x-3x=3x-3x-10\\\\-x=-10\qquad\text{change the signs}\\\\x=10\\\\m\angle1=2(10)+20=20+20=40

4 0
3 years ago
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