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That is wrong

can someone please give me a right answer?

By taking case, the number of possible sequences is 12*11 + 12 + 1 = 145.

Alright. How many total possiblities are there for a dice rolled twice? 36 choices. (6+5+4+3+2+1)

The amount of possiblities for rolling a 1,2, or 3 

1:

only 1,1 works

2:

1,2 or 2,1 works

3:

1,3 or 3,1 works

total of 5 possible choices out of 36

Your answer is 5/36

Please help

By my calculations, the answer is   

1/4*2*(3/13*1 + 1/13*(2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) + 9/13(20)) = 9.12 dollars

triangle BDI is a right triangle (sides are 3-4-5)
That means that triangle ABC is isosceles, since the angle bisector AD is also an altitude.
That means that
BC = 2*BD = 8
AD is also a median of ABC, so AI = 2*DI = 6
The area of ABC is thus 6*8/2 = 24

QR = 120.

See the answer here:   https://web2.0calc.com/questions/algebra_54532

Let's call the perpendicular bisector of PQ as line segment ZY. Since M is the midpoint of PQ, it follows that PZ = ZQ = 18.

We can now use the Pythagorean theorem to find the length of PX and QX:

PX^2 + MY^2 = PQ^2 (since PX is the angle bisector)

PX^2 + 8^2 = 36^2

PX^2 = 36^2 - 8^2

PX^2 = 1296

PX = 36

Similarly,

QX^2 + MY^2 = QR^2

QX^2 + 8^2 = 22^2

QX^2 = 22^2 - 8^2

QX^2 = 484

QX = 22

Since PX is an angle bisector, it follows that PR/PX = QR/QX

PR/PX = 22/36

QR/QX = 36/22

QR = 22 * (36/22) = 36

Since QX + PX = QR = 36, it follows that PX = 36 - QX = 36 - 22 = 14

Finally, we can use the formula for the area of a triangle:

Area = (1/2) * PY * RX

Area = (1/2) * 8 * 14

Area = 56

Therefore, the area of triangle PYR is 56.

The expected value of a single draw is equal to the sum of the probability of each outcome multiplied by the value of the winnings for that outcome.

Let's calculate the expected value:

Ace: 4/52 * 1 = 0.153846

2 through 10: 40/52 * (2 + 3 + ... + 10) / 10 = 1.53846

Face card: 4/52 * 20 = 1.53846

Club: 13/52 * (1 + 2 + ... + 20) / 10 = 2.692308

Spade: 13/52 * (1 * 3 + 2 * 3 + ... + 20 * 3) / 10 = 4.038462

Expected value = 0.153846 + 1.53846 + 1.53846 + 2.692308 + 4.038462, which rounds to 9.19.

The stratagy is a good idea but the exucution is incorrect

Thank you even so

The expected length is 8/5.

1/4 is still incorrect

To find the probability that x>2y, we need to find the area of the region where x>2y in the given rectangular region and divide it by the total area of the rectangular region.

The equation of the line x=2y is a line with a slope of 1/2 and y-intercept 0. The line cuts the x-axis at (0,0) and the y-axis at (2008,1004). Hence, the points of intersection with the given rectangular region are (0,0) and (2008,1004). The region where x>2y is a triangle with vertices at (0,0), (2008,0) and (2008,1004).

The area of the triangle can be calculated as (1/2)baseheight = (1/2)20081004 = 1004*1004.

The total area of the rectangular region is 2008 * 2009 = 4022416.

The probability that x>2y is the ratio of the area of the triangle to the total area of the rectangular region.

Hence, the required probability is:

1004 * 1004 / 4022416 = 1/4022416.

If we try one of those points, lets say (2, 500) 2 is definatly not more than double 500.

We should draw a line from (0,2009/2) to (2008,0). Anything beneath this line is x>2y. It is a traingle so its area is  It should be put over total outcomes equal to 2008 x 2009 = 4,034,072. So we have a probability of 1,008,518\4,034,072 = 1/4.

try listing out possible sequences

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