now.. notice the template above
so hmm
on a) D is 2
on b) D is -2
on c) A is 2
on d) A is 1/2
Five thousand tickets are sold at $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded a
1 answer:
Answer:
the expected value of this raffle if you buy 1 ticket = -0.65
Step-by-step explanation:
Given that :
Five thousand tickets are sold at $1 each for a charity raffle
Tickets are to be drawn at random and monetary prizes awarded as follows: 1 prize of $500, 3 prizes of $300, 5 prizes of $50, and 20 prizes of $5.
Thus; the amount and the corresponding probability can be computed as:
Amount Probability
$500 -$1 = $499 1/5000
$300 -$1 = $299 3/5000
$50 - $1 = $49 5/5000
$5 - $1 = $4 20/5000
-$1 1- 29/5000 = 4971/5000
The expected value of the raffle if 1 ticket is being bought is as follows:
Thus; the expected value of this raffle if you buy 1 ticket = -0.65
You might be interested in
now.. notice the template above
so hmm
on a) D is 2
on b) D is -2
on c) A is 2
on d) A is 1/2
The sum of the series is 44.
Step-by-step explanation:
The given series is
To find the sum of the series, we need to substitute the values for k in the series.
Now, simplifying the square terms, we get,
Multiplying the terms,
Subtracting the values within the bracket term, we get,
Now, adding all the terms, we get the sum of the series,
Thus, the sum of the series is
Answer:
x = 0; y = -6; z = 1
Step-by-step explanation:
You want to find the solution to the system of equations ...
- 9x+y-3z=-9
- 10x-y+2z=8
- -10x-y+4z=10
This set of equations is conveniently solved using the matrix row-reduction features of a scientific or graphing calculator, spreadsheet, or any of a number of apps or on-line calculators.
The attachment shows the solution to be ...
(x, y, z) = (0, -6, 1)
__
<em>Additional comment</em>
Solving a system of three or more equations "by hand" often can be done by an ad hoc process. It can be done in systematic fashion using Gauss-Jordan reduction techniques, but that often gets messy. Similarly, Cramer's Rule can be used, but that math tends to involve more arithmetic operations than are really necessary.
Adding the first equation to the other two eliminates the y-variable and reduces the system to two equations in two unknowns.
(9x +y -3z) +(10x -y +2z) = (-9) +(8) . . . . adding [1] and [2]
19x -z = -1 . . . . simplified
(9x +y -3z) +(-10x -y +4z) = (-9) +(10) . . . . adding [1] and [3]
-x +z = 1 . . . . simplified
At this point, we could graph the two equations, or we can proceed algebraically.
Adding these two equations gives ...
(19x -z) +(-x +z) = (-1) +(1)
18x = 0 ⇒ x = 0
Using the second of the reduced equations, we find ...
-x +z = 1
0 +z = 1 ⇒ z = 1
And using the second of the original equations, gives us ...
10(0) -y +2(1) = 8
-y = 6 . . . . . subtract 2
y = -6 . . . . multiply by -1
Then the solution is (x, y, z) = (0, -6, 1), as above.
