Woovis danced with 3/4 of the 24 dogs at the dance. How many dogs did Woovis dance with?
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The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.
Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.
It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.
Also, let be the quantity of nonzero elements in
Let .
Parity demands that and
must equal 2 or 4.
Case 1: =4 and
=4. There are
=3 ways to put 2-1's in
, so there are 3 ways.
Case 2: =2 and
=4. There are 3 ways to position the -1 in
, 2 ways to put the remaining -1 in
(just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of
for a complete of 24 ways.
Case 3: . There are 3 ways to put the -1 in
. Now, there are two cases on what happens next.
- The 1 in
goes directly under the -1 in
. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in
and
. (Either the 1 comes first in
- The 1 in
Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192
Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.
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PART A
The barn is constructed by the following 2D shapes
Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80
Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600
Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350
One base of length = 45' and width = 15'
Area = 45 × 15 = 675
Two rectangles on each on one side of the roof. We have the length = 45' but not the width. We can work out the width by using Pythagoras theorem
w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77
Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33
Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)
PART B
Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans
PART C
Total cost of paint = 65 × 23.50 = $1527.50
PART D
The barn is constructed by a cuboid and a rectangular prism
V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500
V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800
Total volume = 1800 + 13500 = 15300
The barn is constructed by the following 2D shapes
Two triangles of height = 4' and base = 20'
Area = 2 × (0.5×4×20) = 80
Two rectangles of length = 20' and width = 15'
Area = 2 × 20 × 15 = 600
Two rectangles of length = 45' and width = 15'
Area = 2 × 45' × 15' = 1350
One base of length = 45' and width = 15'
Area = 45 × 15 = 675
Two rectangles on each on one side of the roof. We have the length = 45' but not the width. We can work out the width by using Pythagoras theorem
w² = 4² + 10²
w² = 16 + 100
w² = 116
w = √116 = 10.77
Area of the two rectangles on the roof part is = 2 ×10.77 × 45 = 969.33
Total area to paint = 969.33+675+1350+600+80 = 3674.33 ≈ 3700 (to the nearest hundreth)
PART B
Numbers of paint cans needed = 3700 ÷ 57 = 64.9 ≈ 65 cans
PART C
Total cost of paint = 65 × 23.50 = $1527.50
PART D
The barn is constructed by a cuboid and a rectangular prism
V of cuboid = length × width × height
V of cuboid = 20 × 45 × 15
V of cuboid = 13500
V of triangular prism = Area of cross section × depth
V = [0.5×4×20] × 45
V = 1800
Total volume = 1800 + 13500 = 15300