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nirvana33 [79]
3 years ago
15

Helllllpppppppppp idk what it issssss

Mathematics
1 answer:
yulyashka [42]3 years ago
8 0
I think it’s either a or c
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Which set of ordered pairs in the form of (x,y) does not represent a function of x? {(-1,2), (3,-2),(0,1),(5,2}
serg [7]
For a relation to be a function it must be one-to-one or many-to-one.

One-to-many relation is not a function.

In the above options, C is One-to-many relation therefore cannot be a function.

The reason is that 3 alone maps onto -2 and 5, in the ordered pairs (3,-2) and (3,5). This disqualifies it from being a function.

Hence the graph of this relation will not pass the vertical line test

The correct answer is C
5 0
3 years ago
What are the mean absolute deviation and population standard deviation of this data set? Round to the nearest tenth if necessary
tekilochka [14]

Answer:

For a data set of N elements:

{x₁, x₂, ..., xₙ}

The mean is:

M = \frac{x_1 + x_2 + ... + x_n}{N}

The mean absolute deviation is:

MAD = \frac{|x_1 - M| + ... + |x_n - M|}{N}

And the population standard deviation is:

PSD = \frac{\sqrt{|x_1 - M|^2 + ... + |x_n - M|^2} }{\sqrt{N}}

In this case, our set has 5 elements, and the set is:

{2, 4, 6, 9, 14}

The mean of this set is:

M = \frac{2 + 4 + 6 + 9 + 14}{5} = 7

The mean absolute deviation is:

MAD = \frac{|2 - 7| + |4 - 7| + |6 - 7|+ |9 - 7| + |14 - 7| }{5} = 3.6

And the population standard deviation is:

PSD =  \sqrt{\frac{|2 - 7|^2 + |4 - 7|^2 + |6 - 7|^2+ |9 - 7|^2 + |14 - 7|^2 }{5}} = 4.2

6 0
2 years ago
Determine if biconditional is true. <br> A) Felipe is a swimmer if and only if he is an athlete.
Fynjy0 [20]
It is not a true biconditional.
6 0
3 years ago
Using the figure below find what m&lt;_7 equals. (show work if possible)
miss Akunina [59]

Answer:

68

Step-by-step explanation:

m7 + m8 = Straight line = 180

So, m7 = 180 - 112 = 68

8 0
2 years ago
How many 3/4-foot lengths of pipe can be cut from a 6 1/3-foot pipe?
Ymorist [56]
8, 3/4-foot lengths of pipe can be cut from a 6 1/3-foot pipe.
Hope this helps you :)
5 0
3 years ago
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