Hello good person. The answer is C, (4,7). If you replace the x's in both equations (4) and the y's i both equations (7) you would get your answer for both.
There are 25 species of trees, each with a known abundances. The question is how many possible ways to randomly select one tree there are.
We should calculate the number of combinations. Combinations, because we select item/s from a collection. In this case, when we select only one item, the combination is also a permutation. From set of n objects we select r. In our case: n=25, r=1.
The equation is: n!/r!(n-r)!= 25!/1!*24!=25*24!/24!=25
There are 25 different outcomes (events).
X = -8.
subtract 19.
-X = 8
divide by -1.
X = -8
Answer:
The answer is
<h2>
</h2>
Step-by-step explanation:
The midpoint M of two endpoints of a line segment can be found by using the formula
<h3>
</h3>
where
(x1 , y1) and (x2 , y2) are the points
From the question the points are
(-3,13) and (10,-4)
The midpoint M is
<h3>
</h3>
We have the final answer as
<h3>
</h3>
Hope this helps you
<h3>
Answer: 29,030,400</h3>
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Explanation:
Let's say the mother will temporarily take the place of all the sisters. Wherever the mother sits will represent the block of girls.
Taking out the 6 sisters and replacing them with the mother leads to 7+1 = 8 people in a line.
There are 8! = 8*7*6*5*4*3*2*1 = 40,320 different ways to arrange 8 people.
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Again, the mother's position is where the sisters block will go. So let's say the mother was in seat #2. This would mean one brother would take seat #1, and all of the sisters would take the next six seats, until we reach seat #7 is when another brother would take the next seat.
Within any given permutation (the 40320 mentioned), there are 6! = 6*5*4*3*2*1 = 720 different ways to arrange just the girls in that girls block/group.
All together, there are 40320*720 = 29,030,400 different ways to arrange the 13 siblings where all the girls are seated together.
