11, 12, 13, 14, 15, 16, 17, 18, 19 Sorry if it doesn't help, I'm being too obvious
Answer:
Step-by-step explanation:
When finding the average of a set of numbers, we
- A) Add up all the numbers
- B) Divide that sum by the total number of numbers in the data set
Let's represent this with a formula, s being the sum of all numbers, n being the total number of numbers.
Since we <u>already know the sum</u> and we <u>know how many numbers there are</u> (8), we can substitute inside our expression to find the average.
Therefore, the average of all these numbers is 11.25.
Hope this helped!
Answer:
K = -72
Step-by-step explanation:
1. Get K/9 by itself to do so add 3 to both sides eliminating it on the left by turning the -11 into -8
2. Multiply by 9 on both sides eliminating the division of 9 on the left and turning -8 into -72 leaving the equation as K = -72
To eliminate a number you must do its opposite so the opposite of -3 is +3 so you +3 to both sides. And always start the equation by eliminating what is not connected to the variable in this case the 9 was connected to the variable as K/9.
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
Answer:
"Commutative property of addition and of multiplication"
Step-by-step explanation:
This is the "commutative property" of addition and of multiplication. That as its name indicates tells us for additions, that if we add A + B, or B + A, we should get the same result. Similarly, the "commutative property" of multiplication tells us that if we perform the product A * B we should get the same answer as if we perform the product: B * A.
