The choice isB
all the numbers in that choice decrease in numerical order
http://lianmath.com/describe-a-serie-of-shifts-that-translates-the-graph-yx-93-4-back-onto-the-graph-of-yx-3/
For the given situation we have a total of 259,459,200 permutations.
<h3>
How many permutations are?</h3>
First, how we know that it is a permutation?
Because the order matters, we aren't only selecting 8 out of the 15 people, but these 8 selected also have an order (is not the same thing to finish the race first than fourth, for example).
Then we need to find the number of permutations, to do it, we need to find the numbers of options for each of the 8 positions.
- For the first position there are 15 options.
- For the second position ther are 14 options (one runner already finished).
- For the third position there are 13 options.
- And so on.
Then the total number of permutations (product between the numbers of options) is:
P = 15*14*13*12*11*10*9*8 = 259,459,200
If you want to learn more about permutations:
brainly.com/question/11732255
#SPJ1
X=2.412 is the correct answer
The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04.
Let \displaystyle PP be the student population and \displaystyle nn be the number of years after 2013. Using the explicit formula for a geometric sequence we get
{P}_{n} =284\cdot {1.04}^{n}P
n
=284⋅1.04
n
We can find the number of years since 2013 by subtracting.
\displaystyle 2020 - 2013=72020−2013=7
We are looking for the population after 7 years. We can substitute 7 for \displaystyle nn to estimate the population in 2020.
\displaystyle {P}_{7}=284\cdot {1.04}^{7}\approx 374P
7
=284⋅1.04
7
≈374
The student population will be about 374 in 2020.
