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Lesechka [4]
3 years ago
13

PLLLLLLLEAAAASSE hELPPPPP i WILL GIVE BRAINLIEST TO THE FIRST PERSON WHO ANSWERS CORRECTY!!!!

Mathematics
1 answer:
iren [92.7K]3 years ago
5 0

Hello there,

The answer will be <u>1</u><u>5</u><u> </u><u>students.</u>

<h3>Step-by-step explanation:-</h3>

We will solve the equation as follows:-

First add the total of the students in band and chorus -

→ 70 + 95 = 165

Then subtract the previously got number with given number of chorus students:-

→ 165 - 150 = 15

So, there are <u>15 students</u> who are there in both band and chorus.

✍️ <em>By </em><em>Benjemin</em> ☺️

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Please help!!!
AlladinOne [14]

it would be 13

Step-by-step explanation:

6 0
3 years ago
Rectangle A has 7 times the area of rectangle
krok68 [10]
Wa=0.5Wb, Ha=20, Hb is unkown
the area of rectangle a is : Wa*20
the area of rectangle b is : Wb*Hb
Rectangle a has 7 times the area of rectangle b: Wa*20=7*Wb*Hb
Replace Wa with 0.5Wb: 0.5Wb*20=7Wb*Hb => Hb=10/7
7 0
3 years ago
One number exceeds another by 44. the sum of the numbers is 24. what are the​ numbers?
baherus [9]

Let, the numbers = x,y

x+y= 24   1st eqn

x-y = 44   2nd eqn

24-y-y = 44

-2y = 44-24

y = 20/-2 = -10

substitute that in equation 1st x = 24+10 = 34

so, the numbers would be 34 & -10



4 0
3 years ago
Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

7 0
3 years ago
There are 40 runners in a race. How many ways can the runners finish​ first, second, and​ third?
k0ka [10]

Answer:

117600

Step-by-step explanation:

50 x 49 x 48 = 117,600 ways

You have 50 possible for 1st, 49 for 2nd, 48 for 3rd

or

Permutations 50P3 = 117,600

5 0
3 years ago
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