Only one way to have to take 3 times to select a red ball. You have to select the two blue balls on the first two draws;
Number of all possible arrangements of draws = 5! / [2! * 3!] = 10
Probability = 1 / 10
Nice, Bosco !!!!
41409 / 9 = 4601
Factorization of 4601 = 43 * 107
sum = AB + CDE = 43 + 107 = 150
E = (10, 10sqrt 3) F = (10sqrt 3, 10)
The area is compsed of a square whose area = EF^2 = [ 2(10 - 10sqrt3)]^2 = 400 (2 -sqrt 3) =
800 - 400sqrt 3
And 4 equal sectors = 4 [ pi /6 * 20^2 - (1/2) (20^2) sqrt (3) /2)] = 4 (200pi/3 - 100) =
(800/3)pi - 400
Total area = 400 - 400sqrt 3 + (800/3)pi = 400 [ 1 - sqrt 3 + 2 pi / 3 ]
Here's what I get
1,2,3,4,5,5,5,5,5,5,5,5,6,7,8,9
Mean = 5
Median = 5
Mode = 5
Largest value = 9
surface area = 2 (LW + LH + WH) = 24 ⇒ LW + LH + WH = 12
sum of edge lengths = 4 [ L + W + H) = 24 ⇒ L + W + H = 6
Using a little logic and assuming integer values for each dimensinon, each edge measures 2
Proof
L + W + H = 6 LW + LH + WH =12
2 + 2 + 2 = 6 2*2 + 2*2 + 2*2 = 12
The volume = LWH = 2*2*2 = 8 cubic units
P
5 8
Q M R
5.5 5.5
Law of Cosines (twice)
[QP^2 - QR^2 - PR^2] /[-2* QR * PR] = cos PRQ
[ 25 - 121 - 64] / [-2 * 11 *8 ] = cos PRQ = 10/11
PM^2 = MR^2 + PR^2 - 2 (MR * PR) *cos (PRQ)
PM^2 = 30.25 + 64 - 2[5.5 * 8]* (10/11)
PM ^2 =14.25
PM =sqrt [14.25] ≈ 3.77
At 60 mph, it takes him 20 minutes to drive 20 miles
After 15 min , he has traveled 15 miles
The time it akes the taxi to arrive is irrelevant
So....he has 85 miles left to travel at 50mph
The total time in the taxi is 85 m / 50 m/h = 1.7 hrs
2
11
Two primes
Either the segment connecting these two points is an edge or it is the diagonal of a square
If it is an edge the area = (12 + 98)^2 + (56 -34)^2 = 12584 units^2
If it is a diagonal the area = (diagonal length) /sqrt (2) ) ^2 = [ sqrt [ 12584] / sqrt 2]^2 = 6292 units^2
Sum of all possible area values = 12584 + 6292 = 18,876 units^2
Powers of 4
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
In general 4^(2n) ends in 6
So...the units digit of 1^1 + 2^2 + 3^4 + 4^8 =
1 + 4 + 1 + 6 = 2
Assuming a right triangle
Let the legs = AC = sqrt 7 and BC = sqrt 3
The hypotenuse =
sqrt [ (sqrt .7)^2 + (sqrt .3)^2] = sqrt [ 7 + 3] = sqrt 10 = AB
So....noting that cos B = sinA
cos^2 B + cos^2 A =
sin^2 A + cos^2A = 1
(sqrt (3/10)^2 +(sqrt (7/10)^2 = 1
3/10 + 7/10 = 1
Then angle C is opposite the hypotenuse so its measure = 90°
And cos C = cos (90°) = 0
Angle DAB =90
So angle DAE = 90 / 3 = 30
And DAE is a 30- 60 -90 right triangle
DE is opposite the 30 degree angle
And AD is opposite the 60 degree angle
So AD = 6sqrt 3 = BC
So
BF = BC - FC = 6sqrt 3 - 1 = 5sqrt 3
semi-circle on top right with radius = 4 = 16pi/2 = 8pi km^2
area of rectangle on left side of semi-circle = 8 * 5 = 40 km^2
area of small rectangle in the middle = 2 * 3 = 6 km^2
area of rectangle at bottom = 5 * 8 = 40 km^2
Total area = [ 8pi + 40 + 6 + 40 ] km^2 =
[ 86 + 8pi] km^2 ≈ 111.13 km^2
A
F
3
D E
x 3
B C
AF / AE = AD / AC ⇒ AE/AC = AF/ AD
AD/ AE = AB / AC ⇒ AE / AC = AD/ AB
AF / AD = AD / AB
AD^2 = AF * AB
We need to know AF to solve this rather than EC
If BAC = 60
Then angle BAD = 30 and angle CAD = 30
So CAD can't equal 45
arcos (sqrt (.7)) + arcos (sqrt(.3)) = 90°
cos C = 180 - 90 = cos (90°) = 0
