MHB Solve the a/b and 10/11 Problem: Understanding Quantity Relationships

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The discussion revolves around comparing the quantities a/b and 10/11 to determine their relationship. The initial approach involved converting the fractions to decimal form, but this did not yield a clear answer. After analyzing the inequalities, it was concluded that the relationship between a/b and 10/11 cannot be determined with the given information. The final consensus is that option D, indicating that the relationship cannot be determined, is the correct answer. Additionally, there were inquiries about using LaTeX for better representation of mathematical expressions.
CharlesLin
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this is a nother problem from my study guide.
it starts with a statement
View attachment 4915

A) a/b

B) 10/11
based on this information you have to choose an answer:

a)quantity A is greater
b)quantity B is greater
c)the two are equal
d)the relationship can not be determined from the information given what I first try was was to change every number into a decimal number. for example 9/10= .9 and 11/12=.916. Then I know that a/b is between .9 and .910 but that did't helped because when I calculated 10/11=.90909 the decimal value is also between .9 and .916. then I'm not sure if the answer is d) the relationship can not be determinde...I suspect this is the answer but I'm not entirely shure.

how would you solve this problem?

I also try to picture the solution drawing the line where every fraction would be but I wasn't able to pictured well. Moreover, I had some problems trying to use LATEX there is any tutorial or guide to use it?
 

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If it would hold that $\frac{a}{b}> \frac{10}{11}$ then it would have to hold $\frac{10}{11}< \frac{11}{12} \Rightarrow 10 \cdot 12< 11 \cdot 11$, which is true.

If it would hold $\frac{a}{b}< \frac{10}{11}$ then it would have to hold that $\frac{9}{10}< \frac{10}{11} \Rightarrow 99<100$, which again is true.It also holds that $\frac{9}{10}< \frac{10}{11}< \frac{11}{12}$.

Thus I think that we cannot determine the relationship between $\frac{a}{b}$ and $\frac{10}{11}$ from the given information.
 
I see now
I feel more confident saying that D) is the answer.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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