93 divided by 20 .................................................

Describe the distribution of the data and if the mean or median would be a better measure of center for each. Include outliers.

Bumek [7]

Answer:

Is there a picture you can take

Step-by-step explanation:

8 0

3 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$

And so the equation is

$x^2 - \dfrac{5}{4} + \dfrac{3}{8} = 0$

In order to have integer coefficients, you can multiply both sides of the equation by 8:

$8x^2 - 10 + 3 = 0$

5 0

3 years ago

Help please, I don’t understand this

Solnce55 [7]

Answer:

r(14) = 29

Step-by-step explanation:

r(x) = 3/2x+8

r(14) = 3/2*14+8 = 21+8 = 29

6 0

3 years ago

Read 2 more answers

Plzzzzzzzzzzz help me

oksano4ka [1.4K]

The second table does not represent a function

6 0

3 years ago

PLEASE HELP WITH GRADE 11 MATH. Make sure you show the formula. Substitute values and show all mathematicall operations! show yo

Degger [83]

3x+y

x

=−3

=−y+3

The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:

\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}

3x+y

3(−y+3)+y

−3y+9+y

−2y

y

=−3

=−12

=6

Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:

\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}

x

=−y+3

=−(6)+3

=−3

The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.

We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.

\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}

3x+y

3(−3)+6

−9+6

−3

=−3

=

?

−3

=

?

−3

=−3

Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:

\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}

7x+10y

−2x+y

=36

=9

In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:

\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}

−2x+y

y

=9

=2x+9

Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:

\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}

7x+10y

7x+10(2x+9)

7x+20x+90

27x+90

3x+10

3x

x

=36

=4

=−6

=−2

Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:

\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}

y

=2x+9

=2(−2)+9

=−4+9

=5

The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.

5 0

2 years ago