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Annette [7]
3 years ago
14

While attending a film festival you decide that there are 13 movies that you are interested in seeing. However, you only have ti

me to see 7 movies. In how many different ways can you watch 7 of the 13 movies?
Mathematics
2 answers:
kifflom [539]3 years ago
5 0
<span>A. 8,648,640
B. 13
C. 665,280
D. 4,324,320
</span>Assume you don't watch movies twice.
If you have 13 movies to pick from, then you have 13 choices for the first movie. Then, only 12 choices are left.  Do this until only 6 are left (13-7=6)
This equates to 13*12*11*10*9*8*7=8,648,640
Therefore, the answer is A, 8,648,640 different ways.

Hope this helps!
Afina-wow [57]3 years ago
4 0

Answer:

8,648,640 is correct.

Step-by-step explanation:

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Answer:

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3 years ago
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Vlada [557]

Answer:

y = x - 5

Step-by-step explanation:

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5 0
3 years ago
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Fynjy0 [20]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
Use the identity (x+y)(x2−xy+y2)=x3+y3 to find the sum of two numbers if the product of the numbers is 10, the sum of the square
professor190 [17]

Answer:

The sum of two numbers is 7.

Step-by-step explanation:

Given : the product of the numbers is 10, the sum of the squares of the numbers is 29, and the sum of the cubes of the numbers is 133.

Using identity (x+y)(x^2-xy+y^2)=x^3+y^3 and given details, we have to find the sum of two numbers.

Since, given that the product of the numbers is 10 that is xy=10

Also, given the sum of the squares of the numbers is 29 that is x^2+y^2=29

and the sum of the cubes of the numbers is 133 that is x^3+y^3=133

Using, the given identity (x+y)(x^2-xy+y^2)=x^3+y^3,

Substitute, the given values, we have,

(x+y)(29-10)=133

Simplify , we get,

(x+y)(19)=133

Divide both side by 19, we have,

(x+y)=7

Thus, the sum of two numbers is 7.

7 0
3 years ago
Read 2 more answers
Is this correct???
GaryK [48]

Answer:

There's supposed to be nothing in the 7 and 8 columns

Step-by-step explanation:

You were good, but there aren't any in the 7 or 8 columns, making 9 an outlier.

hope this helps:)

5 0
3 years ago
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