same as before, is the proportion of one, the same as the other? let's do the same here without much fuss.

![\bf \stackrel{mixed}{1\frac{1}{2}}\implies \cfrac{1\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{3}{2}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{~~2\frac{1}{3}~~}{\frac{3}{4}}=\cfrac{~~4\frac{2}{3}~~}{1\frac{1}{2}}\implies \cfrac{~~\frac{7}{3}~~}{\frac{3}{4}}=\cfrac{~~\frac{14}{3}~~}{\frac{3}{2}}\implies \cfrac{7}{3}\cdot \cfrac{4}{3}=\cfrac{14}{3}\cdot \cfrac{2}{3}\implies \cfrac{28}{9}=\cfrac{28}{9}~~\textit{\Large \checkmark}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B3%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B~~2%5Cfrac%7B1%7D%7B3%7D~~%7D%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Ccfrac%7B~~4%5Cfrac%7B2%7D%7B3%7D~~%7D%7B1%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B7%7D%7B3%7D~~%7D%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Ccfrac%7B~~%5Cfrac%7B14%7D%7B3%7D~~%7D%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B7%7D%7B3%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B3%7D%3D%5Ccfrac%7B14%7D%7B3%7D%5Ccdot%20%5Ccfrac%7B2%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B28%7D%7B9%7D%3D%5Ccfrac%7B28%7D%7B9%7D~~%5Ctextit%7B%5CLarge%20%5Ccheckmark%7D)
<span>1. 3a²-a = a(3a - 1)
2. 7ab³-14b = 7b(ab^2 - 2)
3. x³-x²+x = x(x^2 - x + 1)
4. 12x³-xy² = x(12x^2 - y^2)
5. 5x-6x²+7x³ = x(5 - 6x + 7x^2)
6. 7a+8ab+9a² = a(7 + 8b + 9a)
7.3x²-3xy+6xy²
</span>= 3x(x - y + 2y^2)
Answer:
(x^2-6)(x+2)(x-2)
Step-by-step explanation:
x^4-10x^2+24
(x^2-6)(x+2)(x-2)
The answers are in bold
The shown relative frequencies are 0.18, 0.19, and 0.21, which is pretty close to 0.19 - 0.20. From there you can tell that they are reasonably close to equal.
That means that the probailities are close to uniform and => <span>A uniform propabililty model IS a good model to represent probabilities related to the numbers generated by Claudia's calculator.
Therefore, the theoretical probability that any one number is chosen is likely 0.2.
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