Answer:
P43=4!(4–3)!=241=24
Step-by-step explanation:
There are four choices you can make for the lead reindeer. For each possible choice, there are then three remaining you can choose to fly second, making 4×3=12 choices for the lead pair. For each possible choice there are two remaining reindeer to take up the back position, making 12×2=24 choices for the team of three.
This type of problem is called a permutation problem, and we write Pnr for the number of ways of choosing r items from n possibilities when the order of the items matters. In this case we are choosing 3 reindeer from 4 possibilities, and the order they appear in the flying line does matter, so the answer we want is P43. The general formula is Pnr=n!(n−r)!. For the answer we are looking for we therefore have:
P43=4!(4–3)!=241=24
Answer:
27
Step-by-step explanation:
In this case we associate in total value from 225 to 100%, and they tell us that foreign currencies represent 88%, therefore, in quantity this would be:
225 * 88/100
198
The rest of the coins would then refer to the other coins I collect, which would be:
225 - 198 = 27
That is, 27 coins are the remaining coins.
Answer:
your answer will be option D !!
Step-by-step explanation:
123 + x = 180
x = 180 -123
x = 57
have a great day and good luck ^_^
Okay, here we have this:
Considering the provided information and grades, we are going to calculate the requested GPA, so we obtain the following:
Then we will substitute in the following formula:
Then:
Finally we obtain that the student's GPA for that term is approximately 3.06.
Answer:
B. |x+6|-3
Step-by-step explanation:
Well we can tell by looking at the graph that the line has a y-intercept of +3,
meaning we can cross out choices A and C.
Now we can graph B and D,
Look at the image below ↓
By looking at the graphed lines we can tell that y = |x+6|-3 is the correct answer.
Wel can also conclude that choice B is correct because of its vertex being at (-6,-3).
<em>Thus,</em>
<em>answer choice B is the correct answer.</em>
<em>Hope this helps :)</em>
