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zmey [24]
3 years ago
8

The amount of money that high school students spend on fast food each month is usually between $50 and $200. However, there are

a few students who do not eat fast food at all. What measure of spread would be most appropriate to measure the amount of money that high school students spend on fast food per month?
Mathematics
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

B) Interquartile range

Step-by-step explanation:

Since there are some students that do not eat fast food at all, their amounts spent monthly would be $0.  These are much different than the rest of the data, and thus are outliers.

Outliers can greatly affect the mean.  This tells us the mean may not be the best measure.

Since the mean is affected by outliers, the standard deviation is not the best measure either.

Adding data points of 0 will also affect the range.  This leaves the interquartile range as the better measure.

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33 per seconds
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3 years ago
Alex drew a scale drawing of an auditorium. He used the scale 3 inches = 2 feet. If the stage is 30 inches in the drawing, how w
White raven [17]

Answer:

20 feet

Step-by-step explanation:

The scale is 3 inches is to 2 feet

As a ratio, it would be:

\dfrac{3}{2}

The question is what is the actual length of the stage if the drawing is 30 inches. All we need to do is find a ratio proportional to the scale ratio:

\dfrac{3inches}{2feet} = \dfrac{30inches}{xfeet}

Now to find the missing number, cross multiply the fractions first:

\dfrac{3}{2}=\dfrac{30}{x}

(3)(x)=(2)(30)

3x=60

Then solve for x:

3x=60

x=\dfrac{60}{3}

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The answer is 20 feet

8 0
3 years ago
Given the side lengths determine if the following triangles exist, if they do classify
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Answer:

no

Step-by-step explanation:

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3 years ago
What is the solution to this system of equations? *<br> 11) x - 3y=9<br> x+3y=-3
Alika [10]
In point form (3,-2)
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8 0
3 years ago
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Represent the following expressions as a power of the number a (a≠0): c ((a2)−2)5÷(a4a)3
s2008m [1.1K]

Answer:

1/a²⁷

Step-by-step explanation:

Given: $ \frac{((a^2)^{-2})^5}{a^4.a^3} $

We know that when the bases are same the powers can be added. i.,e.,

                                           (xᵃ)ᵇ = x ⁽ᵃ ⁺ ᵇ⁾

⇒ $ \frac{((a^2)^{-2})^5}{a^4.a^3} \implies \frac{(a^2)^{-10}}{a^7} $

$  \implies \frac{a^{-20}}{a^7} = a^{-20 - 7} = a^{-27} $

Also, $ \frac{x^a}{x^b} = x^{a - b} $

This is nothing but, $ \frac{1}{a^{27}} $.

Note that $ a $ cannot be zero here. The condition is provided in the question as well.

7 0
3 years ago
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