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notsponge [240]
4 years ago
11

Michael drove 210 miles in 3 1/2 hours. Jordan drove 330 miles in 6 hours. Which is an accurate comparison of the rates at whic

h the two people drove?
Mathematics
2 answers:
lys-0071 [83]4 years ago
5 0

Michael = 210 / 3.5 = 60 miles per hour

Jordan = 330/ 6 =55 miles per hour


Michael drove 5 miles per hour faster than Jordan

eimsori [14]4 years ago
3 0

Answer:

M 5

Step-by-step explanation:

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Answer:

\text{C. }60

Step-by-step explanation:

<u>Question from image</u>:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the <em>x</em> marks where it cannot be.

\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}

Let's start with the first position:

\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in 3!=6 ways. Therefore, there are 4\cdot 6=\boxed{24} possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in 3!=6 ways. Therefore, there are 3\cdot 6=\boxed{18} possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are 3!=6 ways to rearrange the remaining 3 digits for a total of 2\cdot 6=\boxed{12} possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still 3!=6 ways to rearrange the remaining three numbers. Therefore, there are 1\cdot 6=\boxed{6} possible numbers when 1 is the fourth digit of the number.

Thus, there are 24+18+12+6=\boxed{60} numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

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