Each participant in a certain study was assigned a sequence of 3 different letters from the set {a, b, c, d, e, f, g, h}. if no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (note, for example, that the sequence a, b, c is different from the sequence c, b,a.) 20

Answers

Answer 1

Answer: Number of possible sequence of 3 in 8 = 8P3 = 8! / (8-3)! = 336
36 were not assigned so number participants = 300 (answer).

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If f(x) = 5x 40, what is f(x) when x = –5?

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Answers

you should time it sorry I have not learn that yet srry

What times what equal negative 60 and differs by negative 11

Answers

You can use the " x-method" to solve this question. You need to find the two factors that multiply to -60 and add/ subtract to -11. Your two factors should be 4 and -15. I hope the picture helps

How many positive intergers less than 25 are divisible by 2, 3, and 5?

Answers

Hello,

All numbers divisible by 2,3,5 are multiple of 2*3*5=30.
The only positive integer multiple of 30 less than 25 is 0.
(I don't know if in US 0 is considerated as positive) in my country yes.

How can you solve -2y+5y=14

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you add the terms that are the same and then simplify.

-2y+5y=3y
3y=14
y=4.666 or 4 2/3

Rewrite this expression using the distributive property then simplify.
2/3(6x+12)

Answers

so distributive property of multiplication says
a(b+c)=(ab)+(ac)

so 2/3(6x+12)=((2/3)6x)+((2/3)12)=12/3x+24/3=4x+8=4(x+2)

2/3(6x+12)

=12/3x+24/3

=4x+8

20. There are 15 members of the show choir. In how many ways can you arrange 4 members in the front row when order does not matter?

Answers

The first chair can be any one of the 15.
   For each of those ...
The second chair can be any one of the remaining 14.
      For each of those ...
The third chair can be any one of the remaining 13.
   For each of those ...
The fourth chair can be any one of the remaining 12.

Number of ways to fill the 4 chairs = (15 x 14 x 13 x 12) = 32,760 .

But ...

Each set of 4 people can be seated in (4 x 3 x 2 x 1) = 24 orders.
So each group of 4 people is represented 24 times among the 32,760.

If the order doesn't matter, you're really asking how many different
groups of 4 people can occupy the front row.

That's (32,760) / (24) = 1,365 sets of 4 members, in any order.

Say that the front four people's names are John, Kathy, Mia, and Alex, in that order. You can arrange them as follows.

John, Kathy, Alex, Mia
John, Alex, Kathy, Mia
John, Alex, Mia, Kathy
John, Mia, Alex, Kathy
John, Mia, Kathy, Alex
Alex, John, Kathy, Mia
Alex, John, Mia, Kathy
There are many more, but my guess would be 16 combinations. Test for yourself if you don't think thats right!

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Can you help please?