F(x) + k - Moves the graph k units up.
k f(x) stretches the graph parallel to y-axis by a facor k
f (kx) stretches the graph by a factor 1/k parallel to x-axis
f(x + k) moves the graph 3 units to the left.
For k negative the first one moves it k units down
for second transform negative does same transfoormation but also reflects the graph in the x axis
For the third transform negative k :- same as above but also reflects in y axis
4th transform - negative k moves graph k units to the right
Answer:
Trinomial
Step-by-step explanation:
This is because there are three terms involved in the equation.
One of them is -7ab^4. Another is 4c^3. The third one is 25.
They don't come out even.
As rounded decimals, the two numbers are
<em>5.54138...</em> and <em>-0.54138...</em>
Answer : yes
explication :
Answer:
Approximately
(
.) (Assume that the choices of the
passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all
floors.)
Step-by-step explanation:
If there is no requirement that no two passengers exit at the same floor, each of these
passenger could choose from any one of the
floors. There would be a total of
unique ways for these
passengers to exit the elevator.
Assume that no two passengers are allowed to exit at the same floor.
The first passenger could choose from any of the
floors.
However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only
floors.
Likewise, the third passenger would have to choose from only
floors.
Thus, under the requirement that no two passenger could exit at the same floor, there would be only
unique ways for these two passengers to exit the elevator.
By the assumption that the choices of the passengers are independent and uniform across the
floors. Each of these
combinations would be equally likely.
Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:
.