Answer:
-1/5
Step-by-step explanation:
(10 ,9) .....(40 , 3)
(9-3)/(10-40)=6/-30=-1/5
What about it? What is your question that ur asking
X is greater than or equal to 1.5
Define three events A,B,C such that
A = first card is a face card
B = second card is a face card (assuming event A happens)
C = third card is a face card (assuming events A & B happen)
No replacements are made
There are 3 face cards (J, K, Q) per suit. There are 4 suits. In total, there are 3*4 = 12 face cards out of 52 total
P(A) = probability that event A happens
P(A) = 12/52
P(A) = 3/13
After event A happens, we have 12-1 = 11 face cards out of 52-1 = 51 total
P(B) = 11/51
After B happens, we have 11-1 = 10 face cards out of 51-1 = 50 left over
P(C) = 10/50
P(C) = 1/5
Based on how the events A,B,C are set up, we can form this equation
P(A and B and C) = P(A)*P(B)*P(C)
P(A and B and C) = (3/13)*(11/51)*(1/5)
P(A and B and C) = 11/1105
P(A and B and C) = 0.00995475113122
Rounded to the nearest tenth, the answer is simply 0.0 which is a lousy answer in my opinion because it implies that the answer is truly 0 instead of something really small. Though to be fair, the result is very close to 0. A much better approach is to round to the nearest hundredth to get 0.01; though I would ask your teacher for clarification and guidance.
note: a probability of 0 means that the event is impossible to happen, yet it is not impossible to pull out three face cards in a row. Is it unlikely? Yes. Because the result is roughly 0.00995 which is a really small decimal close to 0. But it's not 0 itself.
Answer:
<h2>a³-b³ = (a-b)(a²+ab+b²)</h2>
Step-by-step explanation:
let the two perfect cubes be a³ and b³. Factring the difference of these two perfect cubes we have;
a³ - b³
First we need to factorize (a-b)³
(a-b)³ = (a-b) (a-b)²
(a-b)³ = (a-b)(a²-2ab+b²)
(a-b)³ = a³-2a²b+ab²-a²b+2ab²-b³
(a-b)³ = a³-b³-2a²b-a²b+ab²+2ab²
(a-b)³ = a³-b³ - 3a²b+3ab²
(a-b)³ = (a³-b³) -3ab(a-b)
Then we will make a³-b³ the subject of the formula from the resultinh equation;
a³-b³ = (a-b)³+ 3ab(a-b)
a³-b³ = a-b{(a-b)²+3ab}
a³-b³ = a-b{a²+b²-2ab+3ab}
a³-b³ = (a-b)(a²+b²+ab)
a³-b³ = (a-b)(a²+ab+b²)
The long division problem that can be used is (a-b)(a²+ab+b²)
