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Mrrafil [7]
3 years ago
7

2 box plots. The number line goes from 175 to 450. For resort A, the whiskers range from 175 to 375, and the box ranges from 250

to 300. A line divides the box at 375. For Resort B, the whiskers range from 200 to 400, and the box ranges from 300 to 375. A line divides the box at 325.
Compare the two box plots and use the drop-down menus to determine where Kevin should plan to go for vacation.

What is the interquartile range for ski Resort A?
What is the interquartile range for ski Resort B?
Mathematics
2 answers:
tekilochka [14]3 years ago
8 0

Answer:

its 275 for 1, and 325 for 2

Step-by-step explanation:

faust18 [17]3 years ago
8 0

Answer:

Step-by-step explanation:

1.275 2.325

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Juan has 20 books to sell. He sells the books for $15 each. The amount of money Juan makes from selling booka is represented by
lana [24]

Answer:

Step-by-step explanation:

Juan has 20 books to sell. He sells the books for $15 each.

The range of the function is the set of all possible values of the dependent variable. The dependent variable here is the amount of money that is made and this amount depends on

the number of books, x sold

The amount of money Juan makes from selling books is represented by a fucntion. f(x)=15x

The maximum amount that can be made from 20 books at a rate of $15 each would be 20×15 = $300

The minimum amount that fan be made is $0 and this is when no book is sold. Let y = f(x). So the range is

0 lesser than or equal to y lesser than or equal to 300

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3 years ago
Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th
Masteriza [31]

Answer:

Let t_i be the time for the ith task.

We know these times have a certain structure:

  • Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5  \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6

  • The second task takes 1 second t_2=1
  • The fourth task takes 10 seconds t_4=10

So, we have the following system of equations:

t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0  \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

  • Subtract row 2 from row 1 \left(R_1=R_1-R_2\right).
  • Subtract row 2 from row 3 \left(R_3=R_3-R_2\right).
  • Add row 3 to row 2 \left(R_2=R_2+R_3\right).
  • Multiply row 3 by −1 \left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right).
  • Add row 4 multiplied by \frac{3}{2} to row 1 \left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right).
  • Subtract row 4 from row 3 \left(R_3=R_3-R_4\right).

Here is the reduced echelon form for the augmented matrix.

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

c) The additional rows are

\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right

and the augmented matrix is

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]

d) To solve the system you must use these row operations.

  • Subtract row 1 from row 6 \left(R_6=R_6-R_1\right).
  • Subtract row 2 from row 6 \left(R_6=R_6-R_2\right).
  • Subtract row 3 from row 6 \left(R_6=R_6-R_3\right).
  • Swap rows 5 and 6.
  • Add row 5 to row 3 \left(R_3=R_3+R_5\right).
  • Multiply row 5 by 2 \left(R_5=\left(2\right)R_5\right).
  • Subtract row 6 multiplied by 1/2 from row 1 \left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right).
  • Add row 6 multiplied by 1/2 to row 3 \left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right).

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]

The solutions are: (t_1,...,t_6)=(5,1,44,10,90,20).

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