MATHEMATICS HIGH SCHOOL

48 pounds are equal to approximately _______ kg. A. 50
B. 96
C. 200
D. 22

Answers

Answer 1

Answer:

Answer:

The answer would be D. 22

Step-by-step explanation:

There are around 0.453592 Kilograms per pound. Multiply 0.453592 × 48.

Your answer would be 21.772416 would would round up to approximately 22 Kilograms.

Answer 2

Answer:

Answer:

C. 200 This is right ttyl.

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Solve 2/3+5/6 and put answer in simplest form

Answers

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Which equation is represented by the graph below?O 2y + x = 10
O y-2x = -5
O -2y = 10x - 4
O 2y = -4x - 10

Answers

It is actually D what are u taking

How many times does 17 go into 65 no decimals please only remainders if there is one.

Answers

Answer: 3.82

Step-by-step explanation: there is a decimal

Do you have a picture

Roberta has a balance of $4350 on a credit card with an APR of 30.2% paying off her balance in which of these lengths of time will result in her paying the least amount of interest?

Answers

Answer:

8 months

Step-by-step explanation:

yurrr

a teacher and 10 students are to be seated along a bench in the bleachers at a basketball game. In how many ways can this be done if the teacher must be seated in the middle and a difficult student must sit to the teachers immediate left?

Answers

Wow !

OK. The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.

   The total number of possible line-ups is

   (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .

But wait ! We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

   (362,880) x (10) = 3,628,800 ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

Answer:

Step-by-step explanation:

362,880

Find the indicated term of the given arithmetic sequence. a1 = 45, d = –6, n = 8a. 309
c. 3
b. –3
d. 87