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Elanso [62]
3 years ago
13

A drive in theater charges $3.50 per car, the drive in already admitted 100 cars, solve an inequality to find the number of cars

the drive in needs to admit to make atleast 500$
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

They need to admit 43 more cars to make at least $500

<u>Solution:</u>

Given, A drive in theater charges $3.50 per car,  

The drive in already admitted 100 cars,  

Then, money collected until now is $ 3.50 x 100 = $ 350  

We have to solve an inequality to find the number of cars the drive in needs to admit to make atleast 500$

Now, let the number of cars need to be admitted be "n"

Then, inequality is:

collected amount + needed amount ≥ $ 500  

$ 350 + $ 3.50 x n ≥ $ 500

350 + 3.5n ≥ 500

3.5n ≥ 500 – 350  

On solving, we get

3.5n ≥ 150  

n ≥ 42.857

As number of cars can’t be fractions, n = 43

Hence, they need to admit 43 more cars to make at least $500

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