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guajiro [1.7K]
3 years ago
7

Which of the following statements about the expected value of a random experiment is true? A. The expected value is the value yo

u will get after one trial. B. The expected value is the average of the values after 100 trials. C. As the number of trials gets larger, the average of the trials approaches the expected value. D. The expected value never actually equals the average of the trials.
Mathematics
2 answers:
Anna71 [15]3 years ago
8 0

Answer:

C. As the number of trials gets larger, the average of the trials approaches the expected value.

Step-by-step explanation:

The statement that is true would be that, as the number of trials gets larger, the average of the trials approaches the expected value. A certain random experiment can have a various number of trials, while the expected value is going to be the average of those trials. The more trials that an experiment has, the more data that can be gathered and the average can be better calculated. This average keeps getting closer and closer to the expected value as more trials are completed and the data keeps repeating certain values.

Tom [10]3 years ago
3 0

Answer: C

Step-by-step explanation: As the number of trials gets larger the average of the trials approaches the expected value

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It means: In the equation of a straight line (when the equation is written as "y = mx + b"), the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis). This useful form of the line equation is sensibly named the "slope-intercept form".

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Solve the system of equations.<br><br><br><br> −2x+5y =−35<br> 7x+2y =25
Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}

This system of equations can be solved in three different ways:

  1. Graphing the equations (method used)
  2. Substituting values into the equations
  3. Eliminating variables from the equations

<u>Graphing the Equations</u>

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is \text{y = mx + b}.

Equation 1 is -2x+5y = -35. We need to isolate y.

\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7

Equation 1 is now y=\frac{2}{5}x-7.

Equation 2 also needs y to be isolated.

\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}

Equation 2 is now y=-\frac{7}{2}x+\frac{25}{2}.

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}

\bullet \ \text{For x = 0,}

\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7

\bullet \ \text{For x = 1,}

\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}

\bullet \ \text{For x = 2,}

\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}

\bullet \ \text{For x = 3,}

\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}

\bullet \ \text{For x = 4,}

\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}

\bullet \ \text{For x = 5,}

\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5

Now, we can place these values in our table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

As we can see in our table, the rate of decrease is -\frac{2}{5}. In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract -\frac{2}{5} from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be y=-\frac{7}{2}x+\frac{25}{2}. Therefore, we just use the same process as before to solve for the values.

\bullet \ \text{For x = 0,}

\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}

\bullet \ \text{For x = 1,}

\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9

\bullet \ \text{For x = 2,}

\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}

\bullet \ \text{For x = 3,}

\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2

\bullet \ \text{For x = 4,}

\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}

\bullet \ \text{For x = 5,}

\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5

And now, we place these values into the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1                  Equation 2

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}                 \begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

Therefore, using this data, we have one solution at (5, -5).

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