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marta [7]
4 years ago
6

Mike purchased 3 DVDs and 14 video games for $203. Nick went to the same store and bought 11 DVDs and 11 video games for $220. H

ow much is each video game and each DVD?
Mathematics
1 answer:
ZanzabumX [31]4 years ago
8 0
Fist add 203 and 220 and you get 423.
Then add all the dvds 11+3 and you get 14 then
Next add all the video games 11+14 and you get 25
next divide 25 from 423 then you get 16.92
then divide the dvds 423 divided by 14 and you get 30.21

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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
What is the domain and range of y+3=2^x
Bad White [126]

Answer: y>-3 and x is all real numbers


5 0
3 years ago
The graph shows the cost of buying beets at a farm stand
antiseptic1488 [7]

Answer:

Ok what's your question? cna you include the graph?

8 0
3 years ago
Write 1/5 as a terminating decimal
borishaifa [10]

Answer: The answer is 0.2

Step-by-step explanation:

1 divided by 5 is 0.2 because 1 goes into 5 0.2 times.

Hope this helps!

5 0
4 years ago
Which is equivalent to √60n^11/256n^4 after it has been simplified completely? Assume n =0
motikmotik

Answer:

\large\boxed{\dfrac{n^3}{8}\sqrt{15n}}

Step-by-step explanation:

\sqrt{\dfrac{60n^{11}}{256n^4}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\=\sqrt{\dfrac{60}{256}n^{11-4}}=\dfrac{\sqrt{60n^7}}{\sqrt{256}}=\dfrac{\sqrt{4n^{6+1}\cdot15}}{16}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{\sqrt{4n^6\cdot15n}}{16}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt{4n^{3\cdot2}}\cdot\sqrt{15n}}{16}\qquad\text{use}\ (a^n)^m=a^{nm}

=\dfrac{\sqrt{4(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{\sqrt{4}\cdot\sqrt{(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{2\!\!\!\!\diagup^1\sqrt{15n}}{16\!\!\!\!\!\diagup_8}=\dfrac{n^3\sqrt{15n}}{8}

5 0
4 years ago
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