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
The answer is 2/9, there are 9 places over all and 2 of them are a 7 and 4.
$190*3=$570
$0.20*600= $120
$570+$120= $690
Total cost of trip is $690
Step-by-step explanation:
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<span>If you would like to know in which step did the student first make
an error, you can find this using the following steps:
y = 4 - 2z
4y = 2 - 4z
________________
-4(y) = -4(4 - 2z)
</span>4y = 2 - 4z<span>
________________
-4y = -16 + 8z ... Step 2
</span><span>4y = 2 - 4z</span><span>
________________
0 = -16 + 8z + 2 - 4z</span>
<span>16 - 2 = 4z</span>
<span>14 = 4z</span>
<span>z = 14/4 = 7/2</span>
<span>
The correct result would be: Student made an error in Step 2.</span>