We have the set: 12 ; ... ; 29 ; ... ; ... ; 35 ; ... ; ... ; 39 ; ... ; 52

If we want to have the maximum mean so the numbers must be largest. So:

+ the 2nd number is 29

+ the 4th and 5th number is 35

+ the 7th and 8th number is 39

+ the 10th number is 52

So their mean is: \(\left(12+29+29+35+35+35+39+39+39+52+52\right):11=36\)

ANSWER: 36

When the 3-digit number plus the 2-digit number, it's easy to see that the hundreds of the 3-digit number is 9 (A) and R = 1; O = 0. From a few try, we soon see that E = 4; T = 3 and S = 7. So the sum of the letters A,T,E,O is: \(9+3+4+0=16\)

ANSWER: 16

In ΔABC, shown here, the measure of angle BCA is 90°, AC = 12 units and BC = 9 units.
If D is a point on hypotenuse AB , such that AD = 5 units, what is the length of
segment CD? Express your answer in simplest radical form.

The total view of the 5 people is: 5.122.890.142 + 7.818.125.465 + 11.789.267.938 + 13.466.208.534 + 14.492.168.647 = 52.688.660.906 (views)

As we see the figure, we'll see there are 17 paths satisfy the question.

There are total: \(10\cdot9:2=45\left(matches\right)\) matches in the tournament so there are 45 wins and 45 loses.

If there are 6 players have 7 or more wins, there are: \(6\cdot7=42\left(wins\right)\) , so there are 42 loses left but there must have 6 more lose (the battles together) so it is unsatisfy

But if there are 5 players then it will satisfy the question.

Hence, the answer is 5 players.

There are all: \(6\cdot6=36\left(ways\right)\) for the two dice to roll.

The possible ways of the two dice are: 6-5 ; 5-4 ; 4-3 ; 3-2 ; 2-1 and reverse the order

So the probability is: \(\dfrac{5\cdot2}{36}=\dfrac{5}{18}\)

We have: 54 * 3 = 162 so the possible number is 162.

We can do this by using Pascal's triangle. The numbers from the tenth row of the triangle (corresponds to the number in the subsets) is:

1    10   45   120   210   252   210   120   45   10   1

(0)  (1)  (2)   (3)   (4)    .................................  (9)  (10)

Because the question asks about odd number of elements so there are:

\(10+120+252+120+10=512\left(subsets\right)\)

ANSWER: 512 subsets

There are: \(\left(8\cdot10^2-4\right)\cdot\left(8\cdot10^6\right)=6368000000=6,368\cdot10^9\) ten-digit telephone numbers are possible.

The common result of the numbers is: 24. (A = 6*4 ; B = 16 + 8 ; C = 36 - 12)

We have: \(84=2\cdot3\cdot14\) is the least distinct set of possible integers.

So their sum is: \(2+3+14=19\)

ANSWER: 19

She draws a picture in: \(240:20=12\left(minutes\right)\)

If she draws three times as fast then she draws a picture in: \(12:3=4\left(minutes\right)\)

Convert: 6 hours = 360 minutes

So the number of pictures she draw is: \(360:4=90\left(pictures\right)\)

ANSWER: 90 pictures

The length above ground of the top of the pole is: \(30:\left(4+1\right)\cdot4=24\left(ft\right)\)

ANSWER: 24 ft

There are: \(6\cdot6\cdot6=216\left(outcomes\right)\) are there for the three values showing on the top faces of the dice.

ANSWER: 216 outcomes

Forming the two figure we'll see a letter H.

The probability that Christoph will get an A on the next two test is: \(0,25\cdot0,25=\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{16}\)

ANSWER: \(\dfrac{1}{16}\)

The original number of milk ball in Mandy's box is: \(5+3+1+3+5+7+9+11+13+15=72\left(milk-balls\right)\)

ANSWER: 72 milk balls

There are: \(2\cdot\left(27-2\right)\cdot\left(27-2\right)\cdot\left(27-2\right)=2\cdot25^3=31250\) arrangements of station are possible.

We have: \(a=GCF\left(72,48\right)=24=2^3\cdot3\)

We have: \(b=GCF\left(108,144\right)=36=2^2\cdot3^2\)

So the least common multiple of a and b is: \(LCM\left(a,b\right)=LCM\left(24,36\right)=2^3\cdot3^2=8\cdot9=72\)

ANSWER: 72