All categories

3 years ago

Find the mode, median, mean, and range for each data set. Round the mean to two decimals places, where necessary.

Mathematics

1 answer:

Verdich [7]3 years ago

3 0

Answer:

The median, mean, and mode of the data set are 5, 4.8, and both 4 and 5 respectively.

Step-by-step explanation:

To find the median of a data set, line up all values from greatest to least and find the number in the center. For example, in the set 1; 2; and 3, 2 would be the median. In this case, the order would be 2, 4, 4, 4, 5, 5, 5, 6, 6, 7. The two numbers in the middle are 5 and 5, their average is 5, so that would be the median. The mean of a data set is the average of all data values in a set. In this case, there are 10 data values, and their total sum is 48, so the mean of this data set is 4.8. The mode of a data set is the value that occurs the most in a set. In this case, 2 occurs once, 4 occurs three times, 5 occurs three times, 6 occurs twice, and 7 occurs once, so the mode of this data set is both 4 and 5.

You might be interested in

Solve the inequality r-(-5)>-2

Jet001 [13]

Answer:

$r>-7$

Step-by-step explanation:

Solving an inequality is similar to solving an equation. To solve for r, isolate it. Whatever you do to one side, do to the other. Remember that a negative and a negative equal a positive, thus one side of the inequality becomes r + 5. Then, subtract both sides by -5 to find the answer.

$r-(-5) >-2\\r+5>-2\\r>-7$

7 0

3 years ago

The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is th

Dafna11 [192]

Answer:

$a + b \geq 30$

$b - a \geq 10$

Step-by-step explanation:

Given

The sum of the two positive integer a and b is at least 30, this means the sum of the two positive integer is 30 or greater than 30, so we write the inequalities as below.

$a + b \geq 30$

The difference of the two integers is at least 10, if b is the greater integer then we subtract integer a from integer b, so we write the inequality as below.

$b - a \geq 10$

Therefore, the following system of inequalities could represent the values of two positive integers a and b.

$a + b \geq 30$

$b - a \geq 10$

6 0

4 years ago

Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o

Ivahew [28]

Answer:

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

5 0

4 years ago

Write the first five terms of the sequence in which the nth term is an = (n+3)!/n+3

Katarina [22]

Answer:

the first five terms are

a1 = 6

a2 = 24

a3 = 60

a4 = 120

a5 = 210

Step-by-step explanation:

the experission used to find the five terms is

an = n3+ 3n2+ 2n

7 0

3 years ago

Mr.george owns 425 acres of land.if he divides the land into half-acre plots,how many plots will he have?

tatiyna

850
All you do is times 425 by 2
425×2
850
hope this helps

6 0

3 years ago