The answer is the letter B
Y = -5
work:
y = |5| - 10
y = 5 - 10
y = -5
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.
-------------------------------
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 :)
the answer is .91 because you have to move the decimal two places to the left
Applying the rule of modulus for given values, the expression will result to 20.
<h3>What is Modulus or Absolute Values</h3>
The modulus of a value for example |a|=a if a is greater than or equal to zero
and also
|a|=-a if a is less than zero
In the expression given; the modulus of -6 written as
|-6|=-(-6) this is because the value is less than zero
The modulus of 2 written as; |2|=2 this is because the value is greater than zero. and;
|The modulus of -14 written as; |-14|= -(-14) since the value is less than zero.
Hence, we can rewrite the expression as;
we deal with multiplication first following the rules of BODMAS
2×-(-6)-3×2+[-(-14)]
2×6-6+14
and thus adding values
12+14-6
and finally we subtract;
26-6
which will results to the solution of 20.
