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Korvikt [17]
3 years ago
9

Calculate the mean, median, and mode of the following set of data. Round to the nearest tenth. 10, 1, 10, 15, 1, 7, 10, 10, 1, 6

, 13
Mathematics
1 answer:
klemol [59]3 years ago
5 0
First order the data: 
1, 1, 1, 6, 7, 10, 10, 10, 10, 13, 15
the mean is all the numbers added together divided by how many there are, 1 + 1 + 1 + 6 + 7 + 10 + 10 + 10 + 10 + 13 + 15 = 84    84/11 = 7.636364 ≈ 7.6
the median is the number in the middle of the list, count numbers from each end, until you either have one or two left, if one left then its that one, if two left then its half way between the two
the median for your set is 10
The mode is the most common number, in your set 10
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Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
What is the lmc of two numbers that have no common factors greater then 1
Luba_88 [7]
We can take prime numbers so they don't have a common factor other than 1.

3, and 5.

So we have to list the multiples to find the LCM

3: 3, 6, 9, 12, 15, 18
5: 5, 10, 15, 20, 25

15 is the LCM of 3 and 5

So, you can notice that if two numbers, lets call them a and b, have no common factor, then a \times b is the LCM.

Hope that helped :)

6 0
4 years ago
Question 3. Solve each equation given<br>below.<br>5 x - 4=31​
harkovskaia [24]

Answer:

x = 7

Step-by-step explanation:

5x-4 = 31

5x = 31+4

5x = 35

x = 7

3 0
3 years ago
Which is the best estimate of the product of 1/5 ⋅ 26 ?
Natasha_Volkova [10]
You would set it up as .2 x 26, which would be 5.2
6 0
3 years ago
Letang Industrial Systems Company (LISC) is trying to decide between two different conveyor belt systems. System A costs $280,00
Vanyuwa [196]

We have to calculate the EAC for both the conveyor belt system.

Solutions :

<u>Equivalent Annual Cost or (EAC) for the SYSTEM-A</u>

$\text{Operating cash flow } = \text{Pre-tax annual operating cost (1-tax rate )} + (\text{depreciation expense} \times $tax rate )

                                $=-855,000(1-0.23)+\left[\left(\frac{280,000}{4}\right)\times 0.23\right]$

                               $= (-85,000 \times 0.77 ) + (70,000 \times 0.23)$

                               = $ 49,350

Year      Annual Cost flow     Present value factor    Present Value of Annual

                                                     at 10%                              cash flow

1                    -49,350                  0.909091                          -44,863.64

2                  -49,350                   0.826446                          -40,785.12

3                   -49,350                  0.751315                            -37,077.39

4                   -49,350                 <u> 0.683013   </u>                       <u>-33,706.71  </u>

     Total                                    $ 3.169865                       $ -156,432.86  

Therefore, Net Present value = present value of the annual cash flow - initial investment.

           = 156,432.86 - 280,000

          = $ 436,432.86 (negative)

Now the EAC or the Equivalent Annual Cost for System A :

$\text{EAC}= \text{Net present value / (PVIFA 10 percent, 4 years)}$

        $=\frac{436,432.86}{3.169865}$

        $= 137,681.86 $ dollar (negative)

$\text{Operating cash flow } = \text{Pre-tax annual operating cost (1-tax rate )} + (\text{depreciation expense} \times $tax rate)

$=79,000(1-0.23)+\left[\left(\frac{360,000}{6}\right) \times 0.23\right]$

$=(-78,000 \times 0.77)+(60,000 \times 0.23)$

= -$ 47,030

Year      Annual Cost flow     Present value factor    Present Value of Annual

                                                     at 10%                              cash flow

1                    -47,030                  0.909091                          -42,754.55

2                  -47,030                   0.826446                          -38,867.77

3                   -47,030                  0.751315                           -35,334.34

4                   -47,030                 0.683013                            -32,122.12

5                  -47,030                  0.620921                            -29,201.93

6                  -47,030                 <u> 0.564474  </u>                        <u>  -26,547.21 </u>  

    Total                                     $ 4.355261                       $  -204,827.91

Net Present Value = Present Value of annual cash inflows – Initial Investment

$= 204,827.91 - 360,000$

= -$ 564,827.91 (negative)

EAC for system B:

Equivalent Annual Cost for system B $=\frac{\text{net present value}}{\text{PVIFA 10 \text percent, 6 years}}$

$=\frac{-564,827.91}{4.355261}$

= -$129,688.66 (negative)

7 0
3 years ago
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