What are the independent and dependent variables in the relationship? 2w-5

Mathematics

1 answer:

vaieri [72.5K]3 years ago

5 0

Answer:

The width is independent, w, and the length is dependent, 2w - 5. If discussing the area then the independent is w and the dependent is Area for A = (2w-5)(w).

Step-by-step explanation:

On a graph called the coordinate plane, there are two axis. The horizontal axis is the x-axis and is known as the independent variable. A great example of an independent variable is time. Time is always represented on the x-axis because time passes by. It does not depend on anything.

The other axis is the y-axis. It is the vertical axis on the graph. It is called the dependent variable because its value depends on x. For example, if you were looking at miles per hour, the number of miles would depend on how many hours you traveled. You have to know the time to find miles. This is a dependent variable.

Here w can be anything and it affects what the length 2w - 5 will be. It determines it because it is part of the expression. The independent variable is w and the dependent variable is the length 2w - 5. If discussing the area then the independent is w and the dependent is Area for A = (2w-5)(w).

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) An instructor gives his class a set of 18 problems with the information that the next quiz will consist of a random selection

RSB [31]

Answer:

The probability the he or she will answer correctly is 1.5%

Step-by-step explanation:

In all, there are 18 problems. In this question, the order of which the problems are sorted for the quiz makes no difference. For example, if the question A of the quiz is P1 and question B P2, and question A P2 and question B P1, it is the same thing.

There are 18 problems and 9 are going to be selected. So, there is going to be a combination of 9 elements from a set of 18 elements.

A combination of n elements from a set of m objects has the following formula:

$C_{(m,n)} = \frac{m!}{n!(m-n)!}$

In this question, m = 18, n = 9. So the total number of possibilities is:

$T_{p} = C_{(18,9)} = \frac{18!}{9!(18-9)!} = 48620$

Now we have to calculate the number of desired outcomes. This number is a combination of 9 elements from a set of 13 elements(13 is the number of problems that the student has figured out how to do).

Now, m = 13, n = 9. The number of desired possibilities is:

$D_{p} = C_{(13,9)} = \frac{13!}{9!(13-9)!} = 715$