Sparkles the Clown makes balloon animals for children at birthday parties. At Sidney's party,

Mathematics

1 answer:

Gwar [14]3 years ago

5 0

Answer: 4 balloons and 3 balloons.

Step-by-step explanation:

Each poodle requires 4 balloons and each giraffe requires 3 balloons.

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Maurinko [17]

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(2, 9)

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3 years ago

Geometry help! Will give BRAINLIEST

disa [49]

Ok, so the sum of the angles of a triangle are 180 degrees.

Add the ratios together: 8+3+9 = 20.

The smallest angle is 3/20 * 180, so it has a measure of 27 degrees.

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The sum of two numbers is 70. The difference of the two numbers is 32. What are the two numbers.

IRISSAK [1]

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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.