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TEA [102]
3 years ago
8

There are 9 students in a class: 7 boys and 2 girls.

Mathematics
1 answer:
Nutka1998 [239]3 years ago
4 0

Answer:

The answer is 2 out of 3

Step-by-step explanation:

Since there are only 2 girls in the class. The probability of 2 girls in the same group is low. The maximum chance of the group being all boys is 2 out of 3. The minimum is 1 out of 3, because the girls can be olin different groups.

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One plus one ididwsoxkscjcjscscscxsc
Gwar [14]

Answer: 2

Step-by-step explanation:

5 0
3 years ago
HELP ASAP PLSSSSSSS
AysviL [449]
It would be B because eaxh section either repeats itself or is close to repeating its previous part
7 0
3 years ago
Which is a factor pair of 72 ?<br> A. 12, 6<br> B. 14, 5<br> C. 23, 4<br> D. 24, 2
sineoko [7]

Answer:

A

Step-by-step explanation:

12x6=72

4 0
3 years ago
Read 2 more answers
I need help urgent this is to hard for me to do
Morgarella [4.7K]

Answer:

( 7 , 4 )

given:

  • 4x + 10y = 68
  • 4x - y = 24

make y the subject in equation 2:

4x - y = 24

-y = 24 - 4x

y = 4x - 24

insert this in equation 1:

4x + 10y = 68

4x + 10(4x - 24) = 68

4x + 40x - 240 = 68

44x = 68 + 240

44x = 308

x = 7

solve for y:

y = 4x - 24

y = 4(7) - 24

y = 4

6 0
2 years ago
Read 2 more answers
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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