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amm1812
3 years ago
9

How many ways can a person select 3 videotapes from a collection of 10 tapes?

Mathematics
1 answer:
lesya [120]3 years ago
6 0
If the 10 videotapes are distinct (different) and the order of selection is immaterial, then there are
C(10,3)=10!/(3!(10-3)!)=10!/(3!7!)=10*9*8/(1*2*3)=120 
ways to select three videotapes out of 10.
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Which statement is true about the end behavior of the graphed function? The graph and choices are attached.
nadezda [96]

The correct answer is A) as the x values go to positive infinity, then the graph goes to negative infinity.

We can tell this by looking at the graph values as x goes up. You'll notice, as we get higher and higher, the graph disappears down. This means it is heading towards negative infinity for the rest of the graph.

7 0
3 years ago
Read 2 more answers
What is 25.7 as a mixed number/fraction?
Kitty [74]
It is 25 7/10 or 257/10
8 0
2 years ago
Use the numbers 1-12 one time each to creat four true statements with addition, multiplication, subtraction and division
laila [671]

If you mean that you can only use each number along the four expressions, it looks impossible to me. Here's the proof.

Multiplications and divisions are the most restrictive operations, because only some triplets will work. In particular, you can choose

(2,3,6),\ (2,4,8),\ (2,5,10),\ (2,6,12),\ (3,4,12)

I'm listing them in triplets because you can use them in many ways. For example, the first triple can be used to generate

2\cdot 3 = 6,\quad 3\cdot 2 = 6,\quad 6\div 3 = 2,\quad 6\div 2 = 3

But it doesn't really matter, in all cases you used numbers 2,3, and 6.

If you look closely, all triplets but the last one involve 2. This means that we must use the last one, because otherwise we would use two triplets with two, and we would have repetitions.

So, we surely have to use (3,4,12), either to write 3\cdot 4=12 or 12\div 4=3.

This means that we can't use 3, 4 and 12 anymore. The only triplet remaining is (2,5,10).

For example, let's say that our multiplication is 2\cdot 5 = 10 and our division is 12\div 4 = 3.

We're left with the following numbers:

1,\ 6,\ 7,\ 8,\ 9,\ 11

From here, we're left with a few choices for the addition: if we choose 1+6=7 we're left with 8,9 and 11. We can't write any subtraction with these numbers.

If we choose 1+7=8, we're left with 6, 9 and 11. We can't write any subtraction with these numbers.

If we choose 1+8=9, we're left with 6, 7 and 11. We can't write any subtraction with these numbers.

To recap, we're forced to use 2,3,4,5,10 and 12 for the multiplication and the division, and the remaining numbers don't allow to write an addiction and a subtraction.

Thus, you can't use all the operations involving numbers 1-12 only once.

7 0
3 years ago
A car travels 24 miles every half hour. how fast is it going?
UNO [17]
Speed = (24*2)/(0.5*2) miles/hr
=48 miles/hr
7 0
3 years ago
How many solution does this have and how do you solve it
Vadim26 [7]

Answer:

Infinite solutions

Step-by-step explanation:

To determine the number of solutions a system has, you need its slope. If the slopes are the same, check if the equations are equivalent or not.

To find the slope, rearrange the equation to isolate "y", which makes it slope-intercept form y = mx + b.

"x" and "y" represent points on the line.

"m" represents the slope, which is how steep a line is.

"b" represents the y-intercept (where the graph hits the y-axis).

Isolate "y" in both equations:

x + 3y = 0

3y = -x        Subtract 'x' from both sides

y = -x/3        Divide both sides by 3.

y = -\frac{1}{3}x

The slope is -1/3.

9y = -3x

y = -3x/9        Divide both sides by 9

y = -\frac{3}{9}x        Reduce the fraction to lowest terms

y = -\frac{1}{3}x

The slope is -1/3, but the two equations are the same.

Since the equations are the same, or <u>equivalent</u>, the two lines are always touching. The solution of a system means where two lines intersect, or meet. Therefore, <u>there are infinite solutions</u>.

5 0
3 years ago
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