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Talja [164]
2 years ago
6

The amount of money that college students spend on rent each month is usually between $300 and $600. However, there are a few st

udents who spend $1,300. What measure of spread would be most appropriate to measure the amount of money that college students spend on rent per month?
A. median
B. range
C. standard deviation
D. interquartile range
Mathematics
2 answers:
Vaselesa [24]2 years ago
8 0

Answer: D. interquartile range

Step-by-step explanation:

Range and interquartile range both measures of the spread of data.

When data has no outlier , then the range is the best to measure the spread of data.

When data has outlier , then the interquartile range is the best to measure the spread of data because range gets affected by outliers  .

Given : The amount of money that college students spend on rent each month is usually between $300 and $600. However, there are a few students who spend $1,300.

$1,300 is an extreme value (outlier) as compare to the given range.

Hence, the measure of spread would be most appropriate to measure the amount of money that college students spend on rent per month = " interquartile range"

Rudiy272 years ago
4 0
<span>The <u>correct answer</u> is:

A) median.

Explanation<span>:

When measuring the spread of this data, the first thing we note is that $1300 is an outlier; it is much, much different from the rest of the data. Therefore it has the potential to skew some measures of spread. For instance, the mean (not in this list) would be affected, since 1300 will raise the mean higher than it should be.

Since 1300 affects the mean, it also affects the standard deviation, which is the average distance each point is from the mean. Therefore standard deviation is not the best measure to use.

The range will be affected by this value as well, since this is the highest value; without it, the highest would be much less, which would make the range much less. Therefore range is not the best measure.

The interquartile range is not affected much by an outlier, but it is not usually a measure of spread that we use to describe values such as this.

We typically use the median to describe these values; since it is in the middle, it is not really affected by any outliers, and it is not a range from lowest to highest, such as the interquartile range (lowest quartile to highest quartile). Therefore median is the best measure.</span></span>
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Answer: 19/60 as decimal is approximately: 0.31667 (Rounded to fifth decimal place)


Explanation: hope this helps
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2 years ago
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Sarah, Natasha and Richard share some sweets in the ratio 4:2:3. Sarah gets 13 more sweets than Richard. How many sweets are the
gayaneshka [121]

Answer:

117 sweets

Step-by-step explanation:

Please see attached picture for full solution. (model method)

S, N and R represents the number of sweets Sarah, Natasha and Richard have respectively.

Alternatively,

Let the number of sweets Sarah have be 4x.

Number of sweets Natasha have= 2x

Number of sweets Richard have= 3x

<em>Sarah gets 13 more sweets than Richard</em>

4x= 3x +13

4x -3x= 13

x= 13

Total number of sweets

= 4x +2x +3x

= 9x <em>(</em><em>simplify</em><em>)</em>

= 9(13) <em>(</em><em>subst</em><em>.</em><em> </em><em>x</em><em>=</em><em>1</em><em>3</em><em>)</em>

= 117 sweets

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2 years ago
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use a graphing calculator or other technology to answer the question which quadratic regression equation best fits the data set
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Answer:

Option C is the correct option.

In other words, the quadratic regression y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\: best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation  y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:  is also attached.

Step-by-step explanation:

x                 y

3                470

4                416

5                403

Analyzing Option A:

Considering the equation

y=32.86\:\left(x\right)^2+379.14\left(x\right)-1369.14\:

From (3, 470), putting x = 3

y=32.86\:\left(3\right)^2+379.14\left(3\right)-1369.14\:

y=64.02

From (4, 470), putting x = 4

y=32.86\:\left(4\right)^2+379.14\left(4\right)-1369.14\:\:

y=673.18

From (5, 403), putting x = 5

y=32.86\:\left(5\right)^2+379.14\left(5\right)-1369.14\:

y=1348.06

Analyzing Option B:

y=32.86\:\left(x\right)^2-379.14\left(x\right)

From (3, 470), putting x = 3

y=32.86\:\left(3\right)^2-379.14\left(3\right)

y=-841.68

From (4, 470), putting x = 4

y=32.86\:\left(4\right)^2-379.14\left(4\right)

\:y=-990.8

From (5, 403), putting x = 5

y=32.86\:\left(5\right)^2-379.14\left(5\right)

\:y=-1074.2

Analyzing Option C:

Considering the equation

y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:

From (3, 470), putting x = 3

y=32.86\:\left(3\right)^2-379.14\left(3\right)+1369.14\:

y=527.46

So, the approximately result is (3, 527)

From (4, 470), putting x = 4

y=32.86\:\left(4\right)^2-379.14\left(4\right)+1369.14\:

y=378.34

So, the approximately result is (4, 378)

From (5, 403), putting x = 5

y=32.86\:\left(5\right)^2-379.14\left(5\right)+1369.14\:\:\:

y=294.94

So, the approximately result is (5, 295)

Analyzing Option D:

Considering the equation

y=-1369.14\:\left(x\right)^2-379.14\left(x\right)+32.86

From (3, 470), putting x = 3

y=-1369.14\:\left(3\right)^2-379.14\left(3\right)+32.86\:\:

y=-13426.82

From (4, 470), putting x = 4

y=-1369.14\:\left(4\right)^2-379.14\left(4\right)+32.86

y=-23389.94

From (5, 403), putting x = 5

y=-1369.14\:\left(5\right)^2-379.14\left(5\right)+32.86\:\:

y=-36091.34

Therefore, from the above calculations and analysis, we conclude that Option C is the correct option.

In other words, the quadratic regression y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\: best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation  y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:  is also attached.

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