<span>Three numbers can be defined as x, y, and z.
x + y + z = 64
y = x+3
z = 2x - 11
.
substitute for y and z
x + (x+3) + (2x-11) = 64
4x -8 = 64
4x = 72
x = 18
.
y = x+3 = 21
z = 2x -11 = 2(18) -11 = 36-11 = 25
.
x + y + z = 18 + 21 + 25 = 64
Correct.
.
Answer: The three numbers are 18, 21, and 25.</span>
Isn't that a binomial problem
Answer:
Step-by-step explanation:
Expand:
-1(x^2-6x+9)+4=-x^2+6x-9 + 4 = -x^2 + 6x - 5
Well, when you think about it a random set of prime numbers would be 1,3,5,7,11,13...
If you add 1+3+5 you get 9 which isn't a prime number.
Another example is 7+11+13 you get 31 which is a prime number.
So to answer your question: it depends on the type of prime numbers that you use
Assuming a standard deck, we started with 52 prior to dealing, then drew 4 spades and 1 non-spade. This leaves 47 cards in the deck. From these 47, we hope to draw another spade. The deck originally started with 13 spades, and after dealing the five cards, the remaining deck contains 9 spades. The probability of drawing one of these 9 cards from the deck of 47 is

That is, we want any 1 spade from the 9 spades available; we don't want any of the other 38 non-spades; and we're drawing 1 card from a total of 47.