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leva [86]
3 years ago
6

PLEASE HELP WITH THIS! if there's is work PLEASE include thank you:) I need # 27 btw

Mathematics
1 answer:
amid [387]3 years ago
8 0
Order.

9, 12, 12, 14, 15, 16, 18, 21

Now split into quarters.

9, 12, 12, 14, 15, 16, 18, 21
          |          |           |
        (1)                   (3)

Determine the values of (1) and (3) by using medians.

14 + 12 + 12 + 9 / 4
47 / 4
approx. 12

So Q1 = 12.

15 + 16 + 18 + 21 / 4
70 / 4
approx. 17

Therefore the answer is C I think. I have not done this ever before. All the knowledge I did was from research lol
You might be interested in
What is the slope pls help
Galina-37 [17]

Answer:

y=.5x+-2.5

Step-by-step explanation:

This is the linear equation (y=mx+b) m means slope so your slope is .5.

7 0
3 years ago
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
I need help asapp...
prohojiy [21]

The value of b is -16 and the value of ac is 60 after comparing with the standard equation.

<h3>What is a quadratic equation?</h3>

Any equation of the form \rm ax^2+bx+c=0 where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}

We have a quadratic function:

= 3x² - 16x + 20

On comparing with standard function:

b = -16

a = 3

c = 20

ac = 3(20) = 60

Thus, the value of b is -16 and the value of ac is 60 after comparing with the standard equation.

Learn more about quadratic equations here:

brainly.com/question/2263981

#SPJ1

3 0
1 year ago
Duc has a trapezoid shaped section of his yard fenced off for his pets. The lengths of the parallel sides of the trapezoid are 8
noname [10]
Area of the trapezoid = 1/2(B+b)h
where
B= length of the longer side of the trapezoid which is equal to 14 ft
b= shorter shorter side of the trapezoid  which equal 8 ft
h = height of the trapezoid which is equal to 4 ft
Area of the trapezoid = 1/2 (14+8)4
Area of the trapezoid yard fence of Duc is 44ft^2
5 0
3 years ago
Read 2 more answers
Using a video camera positioned on a table, Kat records her hanging flower basket to find out what is eating the blooms. The cam
Serga [27]

Answer: 75.5

I took the test and got it right.

8 0
3 years ago
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