is
so it should be between 0 and 1 but a little bit closer to 1.
<span><span>−1/</span>3</span>m−7=5Add 7 to both sides.<span><span><span><span><span><span><span>−1/</span>3</span></span>m</span>−7</span>+7</span>=<span>5+<span>7</span></span></span><span><span><span>
<span><span>−1/</span>3</span></span>m</span>=12</span>Multiply both sides by 3/(-1).<span><span><span>(<span><span><span>3/<span>−1</span></span></span></span>)</span>*<span>(<span><span><span><span>−1/</span>3</span></span>m</span>)</span></span>=<span><span>(<span><span><span>3/<span>−1</span></span></span></span>)</span>*<span>(12)</span></span></span><span>m=<span>−36</span></span>Answer:<span>m=<span>−36</span></span>
There are 14 chairs and 8 people to be seated. But among the 8. three will be seated together:
So 5 people and (3) could be considered as 6 entities:
Since the order matters, we have to use permutation:
¹⁴P₆ = (14!)/(14-6)! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆ x 3!
2,162,160 x 6 = 12,972,960 ways.
Another way to solve this problem is as follow:
5 + (3) people are considered (for the time being) as 6 entities:
The 1st has a choice among 14 ways
The 2nd has a choice among 13 ways
The 3rd has a choice among 12 ways
The 4th has a choice among 11 ways
The 5th has a choice among 10 ways
The 6th has a choice among 9ways
So far there are 14x13x12x11x10x9 = 2,162,160 ways
But the 3 (that formed one group) could seat among themselves in 3!
or 6 ways:
Total number of permutation = 2,162,160 x 6 = 12,972,960
Answer:
1-11 2-4 3-14 4-9 5-8 6-12 7-15
Step-by-step explanation:plz mark brainliest
Answer:
can you be my new master?
Step-by-step explanation: