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AnnZ [28]
2 years ago
15

36 fifth graders are on a team with the same number on students on each team, if there are more then then 1 but less then 20 tea

ms, how many students can be on each team?
Mathematics
1 answer:
frosja888 [35]2 years ago
7 0

Depending on the number of teams, the possible numbers of students per team are in the set:

{2, 3, 4, 6, 9, 12, 18}

<h3>How many students can be on each team?</h3>

We know that there are 36 students divided equally into N teams.

Such that the number of teams is larger than 1 and smaller than 20.

Then we could have 2 teams, such that the number of students in each team is given by the quotient between the total number of students and the number of teams.

36/2 = 18

There are 18 students in each team.

Notice that the numbers of teams can only be factors of 36, where:

36 = 6*6 = 2*2*3*3

So the factors are:

2, 3, 2*3, 3*3, etc...

If there are 3 teams we have:

36/3 = 12 students per team.

If there are 6 teams we have:

36/6 = 6 students per team.

if there are 9 teams:

36/9 = 4 students per team.

If there are 12 teams:

36/12 = 3 students per team.

If there are 18 teams:

36/18 = 2 students per team.

These are all the possibilities.

If you want to learn more about factors:

brainly.com/question/219464

#SPJ1

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Use the figure below to find the lateral surface area. Help needed ASAP. (WILL MARK BRAINLIEST)
o-na [289]

Answer:

60 is your answer

Step-by-step explanation:

Good Luck!!!   :D

Can i have brainliest?

4 0
3 years ago
Please people i will give you lots of points and please give me the correct answer
cupoosta [38]

Answer:

15

Step-by-step explanation:

30 subtract 15 = 15 and both the 10 and the other 10 are the same

5 0
3 years ago
If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

31 and 41

8 0
1 year ago
50 POINTS!!!
r-ruslan [8.4K]

housing: 900/4500 = 0.2 = 20%

transportation: 650 / 4500 = 0.14 = 14%

utilities: 120/4500 = 0.026 = 2.6% ( round to 3% maybe)

food: 475/4500 = 0.105 = 10.5% (round to 11% maybe)

savings: 1600 / 4500 = 0.355 = 35.5% (round to 36% maybe)

Student Loan: 270 / 4500 = 0.06 = 6%

Other expenses: 485 / 4500 = 0.107 = 10.7% (round to 11% maybe)



 You didn't say if they needed to be rounded so I gave you the options on the ones that may need it.

8 0
3 years ago
Need help....... thx
Blababa [14]

I got 75582.

Explanation:

First, group the 40 identical candies into 20 pairs. It doesn't matter how since the candies are identical. This grouping will ensure that any assigment will contain at least two candies.

Then think of the 20 groups a 20 beads on a string. We are looking to place 11 separators between them to obtain 12 segments, each with a varying number of beads between them. How many ways are there to place 11 separators to 19 potential spaces between beads? The asnwer is n = {19\choose{11}}=75582


5 0
3 years ago
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