Answer:
B. 32
Step-by-step explanation:
There are 8 fl. oz. in one cup
8*4=32
Usually one will differentiate the function to find the minimum/maximum point, but in this case differentiating yields:
which contains multiple solution if one tries to solve for x when the differentiated form is 0.
I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.
If, however, the function is
Then differentiating and equating to 0 yields:
which gives:
or
We reject x=5 as it is when it ix the maximum and thus,
, for
Answer: y = f(m)
Justification:
By definition, the graph of a function in the cartesian coordinate plane is the plot of the curve that join all the points that satisfy the equation of the function.
In the cartesian coordinate plane, each point is represented by a pair of coordinates (y, x). y is the value of the function f(x), and x is any input value of the function.
In this case, the function is given as f(m), which means that the input value is represented by m.
Hence, the graph is the plot of y = f(m).
The choice of the letter used for the variable is completely arbitrary and does not change either the math or the graph. It could have been said that the function is f(x) in which case the graph would be y = f(x). Nothing changes but the name that you use.
For there to be 1 car, we consider two possible outcomes:
The first door opened has a car or the second door opened has a car.
P(1 car) = 2/6 x 4/5 + 4/6 x 2/5
P(1 car) = 8/15
For there to be no car in either door
P(no car) = 4/6 x 3/5
P(no car) = 2/5
Probability of at least one car is the sum of the probability of one car and probability of two cars:
P(2 cars) = 2/6 x 1/5
= 1/15
P(1 car) + P(2 cars) = 8/15 + 1/15
= 3/5
The calendar obviously has an integral number of years and months in 400 years. If it has an integral number of weeks, then it will repeat itself after that time. The rules of the calendar eliminate a leap year in 3 out of the four century years, so there are 97 leap years in 400 years. The number of excess days of the week in 400 years can be found by ...
(303·365) mod 7 + (97·366) mod 7 = (2·1 + 6·2) mod 7 = 14 mod 7 = 0
Thus, there are also an integral number of weeks in 400 years.
The first day of the week is the same at the start of every 400-year interval, so the calendar repeats every 400 years.
