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3 years ago

What was the average monthly temperature for the month with the smallest coat sales?

Mathematics

1 answer:

scoundrel [369]3 years ago

6 0

Answer:

Try 23*C

Step-by-step explanation:

The lowest sales month only sold $410 of coats, and the the temp in that month was 23*C

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For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

8 0

3 years ago

the ratio of the number of adults to the number of students at the problem has to be 1 :10 Last year there were 477 more student

JulsSmile [24]

About 48 adults. I think.

4 0

3 years ago

Andrew jogs for 3 miles each day on the weekdays and for 6.5 miles each day over the weekend. What is the average distance that

kirill [66]

Set up the data:
3,3,3,3,3,6.5,6.5
15+13=28
28/7=4

He runs an average of 4 miles a day

7 0

3 years ago

Read 2 more answers

Abdul has 15 shirts to fold. Let n be the number of shirts he would have left to fold after folding f of them. Write an equation

andrew-mc [135]

Answer:

n + f = 15

Step-by-step explanation:

n + f = 15

n + (6) = 15

n = 15 - 6 (opposite side opposite sign)

n = 9

4 0

2 years ago

Raise 12/8 = ?/48 to higher terms

allochka39001 [22]

Let ?=x
12/8= x/48
8x=576
x= 576/8
x= 72
Therefore the higher term fraction is 72/48

6 0

4 years ago