Answer:
210
Step-by-step explanation:
Here comes the problem from Combination.
We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.
Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7
Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6
Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5
Hence the total number of ways for three students to be seated will be
7 x 6 x 5
=210
We call x the 4 points-worth problems and y the 3 points-worth problems
You know that x+y = 32
4x+3y = 111
You know that the difference from the x and y is 32 so write:
x = 32-y
Substitute at x the value of 32-y
4(32-y)+3y = 111
128-4y +3y = 111
-4y+3y = 111-128
-y = -17
y = 17
Answer:
- y = 81-x
- the domain of P(x) is [0, 81]
- P is maximized at (x, y) = (54, 27)
Step-by-step explanation:
<u>Given</u>
- x plus y equals 81
- x and y are non-negative
<u>Find</u>
- P equals x squared y is maximized
<u>Solution</u>
a. Solve x plus y equals 81 for y.
y equals 81 minus x
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b. Substitute the result from part a into the equation P equals x squared y for the variable that is to be maximized.
P equals x squared left parenthesis 81 minus x right parenthesis
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c. Find the domain of the function P found in part b.
left bracket 0 comma 81 right bracket
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d. Find dP/dx. Solve the equation dP/dx = 0.
P = 81x² -x³
dP/dx = 162x -3x² = 3x(54 -x) = 0
The zero product rule tells us the solutions to this equation are x=0 and x=54, the values of x that make the factors be zero. x=0 is an extraneous solution for this problem so ...
P is maximized at (x, y) = (54, 27).
