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Delvig [45]
3 years ago
6

When the tortoise raced the hare, the former maintained a constant pace of one mile per hour throughout the race; while the latt

er, being overconfident, wasted much time and averaged only 3/4 miles per hour for the first half of the course. How fast must the hare run over the second half of the course in order to win?
Mathematics
1 answer:
katen-ka-za [31]3 years ago
6 0
<span>The tortoise is runing one mile per hour the entire race. The hare is runing 3/4 only the first half. If we multiply by 2 totis =2 miles and hare =1.50 the hare need to run 1/2 or .50 faster to beat the tortis.</span>
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Answer:

60%

Step-by-step explanation:

You can solve this problem by setting up a system of equations.

Let's say that the number of tickets bought by students in the first year is x, and the number bought by continuing students is y. From there, you can set it up like this:

0.4x+0.2y=160

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Now, you can multiply the first equation by 5 on both sides to get:

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Subtracting the second equation from the first equation now yields:

x=300

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Since 300 of the 500 tickets bought were from the first year students, and 300/500 is 0.6, 60% of the students who bought the ticket were first year students. Hope this helps!

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