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damaskus [11]
3 years ago
8

What is the difference between discrete vs continuous data

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0
With continuous data, it is possible to find the midpoint of any two distinct values. For instance, if h = height of tree, then its possible to find the middle height of h = 10 and h = 7 (which in this case is h = 8.5)

On the other hand, discrete data can't be treated the same way (eg: if n = number of people, then there is no midpoint between n = 3 and n = 4). 

-------------------------------------

With that in mind, we have the following answers

1) Continuous data. Time values are always continuous. Any two distinct time values can be averaged to find the midpoint

2) Continuous data. Like time values, temperatures can be averaged as well. 

3) Discrete data. Place locations in a race or competition are finite and we can't have midpoints. We can't have a midpoint between 9th and 10th place for instance.

4) Continuous data. We can find the midpoint and it makes sense to do so when it comes to speeds. 

5) Discrete data. This is a finite number and countable. We cannot have 20.5 freshman for instance.
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Factories 54m3n + 81m4n2 b) 15x2y3z + 25x3y2z + 35x2y2z​
Oliga [24]

<u>Part a)</u>

Given the expression

54m^3n\:+\:81m^4n^2

Apply exponent rule:    a^{b+c}=a^ba^c

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn     ∵ m^4n^2=m^3mnn

Rewrite 81 as 3 · 27

Rewrite 54 as 2 · 27

                             =2\cdot \:27m^3n+3\cdot \:27m^3mnn

Factor out the common term:   27m³n

                              =27m^3n\left(2+3mn\right)      

Therefore,

54m^3n\:+\:81m^4n^2=54m^3n+81m^3mnn=27m^3n\left(2+3mn\right)

<u>Part B)</u>

Given the expression

15x^2y^3z+25x^3y^2z+35x^2y^2z

Apply exponent rule:    a^{b+c}=a^ba^c

15x^2y^3z+25x^3y^2z+35x^2y^2z=15x^2y^2yz+25x^2xy^2z+35x^2y^2z

Rewrite as

                                               =3\cdot \:5y^2x^2zy+5\cdot \:5y^2x^2zx+7\cdot \:5y^2x^2z

Factor out common term 5y²x²z

                                                =5y^2x^2z\left(3y+5x+7\right)

Therefore,

15x^2y^3z+25x^3y^2z+35x^2y^2z=5y^2x^2z\left(3y+5x+7\right)

8 0
3 years ago
What is the solution of this equation? 3a = 18
Triss [41]
3a =18

3a/3 = 18/3

a = 18/3

a = 6

Hope this helps
6 0
3 years ago
Read 2 more answers
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