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).
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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.
1/12 if each Question is A-D
You need to factor the numerator and denominator...
(2x^2+4x-2x-4)/(2x^2-2x-2x+2)
(2x(x+2)-2(x+2))/(2x(x-1)-2(x-1))
((2x-2)(x+2))/((2x-2)(x-1)) so the (2x-2)s cancel out leaving
(x+2)/(x-1)
The answer is 48. 98 - 50 = 48 apples left :)
Answer:
Step-by-step explanation:
q is TFTF
~q use negation, not q so is the opposite of q : FTFT
p↔~q use biconditional ,and will be True only is both statements are T or both are F
p values are TTFF ↔~q values are FTFT : FTTF
(p↔~q )∧~q use conjunction, that is True only if both statements are T
(p↔~q ) values are FTTF ∧~q values are FTFT : FTFF
(p↔~q )∧~q → p use a conditional statement, where only True False will give a F
(p↔~q )∧~q values are FTFF → p values are TTFF : TTTT
