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4 years ago

How to rewrite an expression using the GCF?

Mathematics

1 answer:

ArbitrLikvidat [17]4 years ago

4 0

GCF is 3. Divide everything by 3 and get 1x + 4

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Our 6 Grade math class was comparing and different lemonade recipes,one used 2 cups of water to 1 scoop of mixture. The other us

Reptile [31]

The one that used two cups. for the one that used 5 cups of water and 3 scoops of mixture would have a more strong taste.

7 0

4 years ago

HELP FAST ILL MAKE BRAINIEST Mr. McClellan compared the weights (in pounds) of pairs of elk antlers dropped at Mount St Helens N

Anika [276]

a.1 Line plot for eight of elk antler pairs at Mount St Helens NVM

The diagram is shown in figure 1.

a.2 Weight of elk antler pairs at Rocky Mountain NP

The diagram is shown in figure 2.

b. Calculate the following for each set of data:

We can found the mean (µ) of a set of data points by adding them up and dividing by the number of data points.

b.1 Purple Mean:

Set: $\{34, 34, 30, 30, 30, 28, 28, 26\} \rightarrow \frac{34+34+30+30+30+28+28+26}{8} \rightarrow \boxed{\mu=30}$

b2. Red Mean:

Set: $\{40, 38, 36, 36, 36, 36, 34, 32\} \rightarrow \frac{40+38+36+36+36+36+34+32}{8} \rightarrow \boxed{\mu=36}$

b.3 Red Median

The Median is the middle of a sorted list of numbers. The median of a finite list of numbers can be found by arranging all the numbers from smallest to greatest. So, arranging the red set we have:

$\{32, 34, 36, 36, 36, 36, 38, 40\}$

Given that the number of elements is pair, the median can be solved as follows:

 $\overline{x}=\frac{x_{(n/2)}+ x_{((n/2)+1)}}{2}$

∴ $n=8$
∴ $x_{(n/2)}=x_{(4)}=36$
∴ $x_{((n/2)+1)}=x_{(5)}=36$

Then:

$\overline{x}=\frac{36+36}{2}=36$

b.4 Purple MAD

The Mean Absolute Deviation (MAD) is when you find the distance of each data point from the mean and then find the mean of those distances. So:

$\{34, 34, 30, 30, 30, 28, 28, 26\}$ has $\mu=30$
$distances=\{4, 4, 0, 0, 0, 2, 2, 4\} \rightarrow \frac{4+4+0+0+0+2+2+4}{8}= \boxed{MAD=2}$

b.5 Red MAD

$\{40, 38, 36, 36, 36, 36, 34, 32\}$ has $\mu=36$
$distances=\{4, 2, 0, 0, 0, 0, 2, 4\} \rightarrow \frac{4+2+0+0+0+0+2+4}{8}= \boxed{MAD=1.5}$

c. Calculate the means-to-MAD ratio for the two areas of collection

c.1 Purple set:

The ratio is the relationship between the mean and the MAD indicating how many times the first number contains the second, so:

$\frac{\mu}{MAD}= \frac{30}{2}=\boxed{15}$

c.2 Red set:

$\frac{\mu}{MAD}= \frac{36}{1.5}=\boxed{24}$

d. What inference can be made about the areas in regard to weight of dropped elk antlers?

MAD tells us how far, on average, all values are from the middle. So, in the example weight of elk antler pairs at Mount St Helens NVM are, on average, 2 away from the middle. On the other hand, weight of elk antler pairs at Rocky Mountain NP are, on average, 1.5 away from the middle. So, we can assure that elk antler pairs at Rocky Mountain NP weighs more than elk antler pairs at Mount St Helens NVM.

3 0

3 years ago

On a coordinate plane, a line goes through (negative 12, negative 2) and (0, negative 4). A point is at (0, 6).

mariarad [96]

Point (-12 , 8) is on the line that passes through (0, 6) and is parallel to the given line ⇒ 1st

Step-by-step explanation:

Parallel lines have:

Same slopes
Different y-intercepts

The formula of the slope of a line which passes through points $(x_{1},y_{1})$ and $(x_{1},y_{1})$ is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

∵ The given line passes through points (-12 , -2) and (0 , -4)

∴ $x_{1}$ = -12 , $x_{2}$ = 0

∴ $y_{1}$ = -2 , $y_{2}$ = -4

- Use the formula of the slope above to find the slope of the given line

∵ $m=\frac{-4-(-2)}{0-(-12)}=\frac{-4+2}{12}=\frac{-2}{12}=\frac{-1}{6}$

∴ The slope of the given line is $\frac{-1}{6}$

∵ The two lines are parallel

∴ Their slopes are equal

∴ The slope of the parallel line = $\frac{-1}{6}$

∵ The parallel line passes through point (0 , 6)

- The form of the linear equation is y = mx + b, where m is the slope

and b is the y-intercept (y when x = 0)

∵ m = $\frac{-1}{6}$ and b = 6

∴ The equation of the parallel line is y = $\frac{-1}{6}$ x + 6

Let us check which point is on the line by substitute the x in the equation by the x-coordinate of each point to find y, if y is equal the y-coordinate of the point, then the point is on the line

Point (-12 , 8)

∵ x = -12 and y = 8

∵ y = $\frac{-1}{6}$ (-12) + 6

∴ y = 2 + 6 = 8

- The value of y is equal the y-coordinate of the point

∴ Point (-12 , 8) is on the line

Point (-12 , 8) is on the line that passes through (0, 6) and is parallel to the given line

Learn more:

You can learn more about the equations of parallel lines in brainly.com/question/9527422

#LearnwithBrainly

4 0

4 years ago

Read 2 more answers

In 1935, Harvard linguist George Zipf I wanted out that the frequency of the Kth most frequent word in a language is roughly pro

madam [21]

Answer:

C. 33%

Step-by-step explanation:

The frequency of the Kth most frequent word in a language is roughly proportional to 1/K of the most frequent word.

So the 2nd most frequent word is roughly proportional to 1/2 of the most frequent word

The 3rd most frequent word is roughly proportional to 1/3 = 0.333 = 0.33 of the most frequent word.

0.333 is rounded to 0.33 = 33%.

If it had been 0.335 or higher, it would be round to 0.34 = 34%.

So the correct answer is:

C. 33%

7 0

3 years ago

determine the quadratic function of f whose graph is given. The vertex is (1,-3) and the y-intercept is -2

LuckyWell [14K]

Hmmm the y-intercept is at -2? what does that mean? well, is where the graph "intercepts" or touches the y-axis, and when that happens, x = 0, so the point is really ( 0 , -2 ).

and we know where the vertex is at. Let's assume a vertical parabola, in which case the squared variable is the "x".

$\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
vertex
\begin{cases}
h=1\\
k=-3
\end{cases}\implies y=a(x-1)^2-3
\\\\\\
\textit{now, we also know that }
\begin{cases}
x=0\\
y=-2
\end{cases}\implies -2=a(0-1)^2-3
\\\\\\
1=a(-1)^2\implies \boxed{1=a}\qquad thus\implies 
\begin{cases}
y=1(x-1)^2-3\\
\textit{or just}\\
y=(x-1)^2-3
\end{cases}$

6 0

4 years ago