Which Doesn't Belong?

Mathematics

2 answers:

klio [65]3 years ago

8 0

I’m assuming 9? There isn’t a question so

Kryger [21]3 years ago

8 0

Step-by-step explanation:

If you are goin for which one isn't prime, then it would be four.

If you are going for which one isnt a factor, there is more then one

If you are going for something else, i have no idea.

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kupik [55]

If you do 25% of 1588 you get 397 LB’s, so the closest one to that would be A.375

4 0

3 years ago

Help pls I'll give 25 points

RideAnS [48]

Answer:

$\text{C. }60$

Step-by-step explanation:

Question from image:

"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 x 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.