Nombre de usuarioAnthrax
Preguntas 2
Respuestas 45


I agree with your rationale for part a).


As for part b), you can apply the same logic of breaking each draw down. For the first two draws, you would have the same probability of \(\frac{2}{9}\) for drawing two black balls. The probability of having one or more white balls at this point is a complement of your desired event of having two black balls, so the probability of having one or more white balls is \(1 - \frac{2}{9} = \frac{7}{9}\). In other words, you will be redrawing \(\frac{7}{9}\) of the time. The probability of drawing two white balls would be the same as drawing two black balls because there are 5 of each, thus P(both white balls) = \(\frac{2}{9}\). Thus, the only other possibility of drawing exactly one white ball and one black ball would be \(1 - \frac{2}{9} - \frac{2}{9} = \frac{5}{9}\)


Say you drew one white ball in your first drawing. This happens \(\frac{5}{9}\) of the time. If you put the white ball back, your probability of drawing a second black ball from the 9 with 4 black balls would be \(\frac{4}{9}\), so this probability is \(\frac{5}{9}*\frac{4}{9} = \frac{20}{81}\).


Say you drew two white balls in your first drawing. This happens \(\frac{2}{9}\) of the time. If you put both balls back, your probability of drawing two black balls would be the same as the initial drawing, since you have 5 black and 5 white in the bag now. Thus, this probability is \(\frac{2}{9}*\frac{2}{9}=\frac{4}{81}\).


Using the sum rule, as these are all disjoint events, and including the probability we chose two black balls on the first draw, we arrive at the probability being \(\frac{2}{9} + \frac{20}{81} + \frac{4}{81} = \frac{14}{27}\). This could be wrong I gotta get to class lmao

7 feb 2020
14 ene 2020
22 may 2019

I guess I'll solve the second question then hahaha


To restate the important distinction in the question, \(g(x)\) is the table and \(f(x)\) is the graph.

Let's consider the validity of each option:


A: "\(f(x) \) and \(g(x)\) have the same y-intercept." For this to be true, we would want \(f(0) = g(0)\). The y-intercept of \(g(x)\) can be found in the table at \(g(0)\), which is equal to 2. Looking at the graph of \(f(x)\), we are unable to see the y-intercept at \(f(0)\), but it is definitely less than -6. Therefore, \(f(0) \neq g(0)\) and this is not true.


B: "\(f(x) \) and \(g(x)\) intersect at two points." For this to be true, we need two points from each graph that are equivalent to each other. In other words, at two \(x\) inputs, \(f(x)\) and \(g(x)\) give the same output. Let's go through each value from the table of \(g(x)\) and see if it matches \(f(x)\).\(g(-2) \neq f(-2), g(0) \neq f(0), g(2)=f(2), g(4) \neq f(2), \)and \(g(6) \neq f(6)\). We can see that \(f(x) \) and \(g(x)\) intersect at one point, but not at two. Therefore, this is not true.


C: "\(f(x)\) is greater than \(g(x)\) for all values of \(x\)." We can quickly find a counterexample for this. At \(x=4\)\(g(4) = 2\) and \(f(4) = 0\). Therefore, \(f(4) \ngtr g(4)\) and this is not true.


D: "\(f(x) \) and \(g(x)\) have a common x-intercept." We learned from looking at B that \(g(2) = f(2)\), wherein both equaled a y-value of \(0\). Therefore, this is true.


Our answer to this question is then the fourth option, or as I have labeled it, D.

17 may 2019