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seropon [69]
4 years ago
5

Six friends, four boys and two girls, went to a movie theater. They wanted to sit in a way so no girl sits on either first or la

st chair. How many such arrangements are possible?
Mathematics
1 answer:
mel-nik [20]4 years ago
8 0

Answer:

120 ways

Step-by-step explanation:

Since no girl wil sit either first or last

Hence, in all irrespective whether boy or girl = (6 - 1)! = 5! = 5 X 4 X 3 X 2 = 120

∴ Number of arrangements = 120 ways

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5556587 x 10^-2

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Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Oliver has 3 times as many rocks in his collection as Katie. Then Oliver collected 75 more rocks and Katie collected 105 more ro
Evgesh-ka [11]

Answer:

The answer to your question is Katie had 15 rocks and Oliver 45 rocks.

Step-by-step explanation:

Conditions

Katie has x amount of rocks

Oliver has 3x amount of rocks

The final amount of rocks

                                    Oliver = 3x + 75

                                     Katie = x + 105

Now, they have the same amount of rocks, then we can equal both equations

                                   3x + 75 = x + 105

Solve for x

                                     3x - x = 105 - 75

                                           2x = 30

                                             x = 30/2

                                             x = 15

Conclusions

At first Katie had 15 rocks and Oliver had 3(15) = 45 rocks

5 0
3 years ago
Read 2 more answers
What is the solution to the inequality shown below?
lesantik [10]

To solve the inequality, you need to isolate/get the variable "p" by itself in the inequality:

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5p + 26 - 26 < 72 - 26

5p < 46      Divide 5 on both sides to get "p" by itself

\frac{5p}{5}

p

p < 9.2       Your answer is A

8 0
4 years ago
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