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melisa1 [442]
3 years ago
15

Selene wants to invite some friends to go with her to see the El Control concert. She will purchase tickets online for $42.50 ea

ch plus a $5 handling fee. If Selene has $250, how many tickets can she buy? Write and solve an inequality to model the situation.
A
42.50x+5≤250

5 tickets

B
42.5x+5≥250

6 tickets

C
42.50x+5≥250

5 tickets

D
42.50x+5≤250

6 tickets
Mathematics
1 answer:
nikdorinn [45]3 years ago
6 0
The price of each ticket is $42.50. So if Selene buys x tickets, the cost of x tickets will be 42.50x.

She has to pay $5 handling fee as well for purchasing the tickets online. So her total cost of the tickets will be:

Total Cost = 42.50x + 5

Selene got $250. This means she can spend a maximum of $250 on the tickets. Maximum of $250 means less than or equal to $250.

So, we can write the spending on the tickets in the form of following in equality:

42.50x + 5 ≤ 250

42.50x ≤ 245

x ≤ 5.76

This shows, Selene can buy 5.76 tickets. Since the number of tickets has to be a whole number we look for the nearest possible number. She cannot buy 6 tickets, as the cost will exceed $250. So the maximum tickets she can buy are 5.

Therefore, the correct answer is option A
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The answer is:

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\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)

(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

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