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Likurg_2 [28]
3 years ago
14

Which statement below is incorrect? The mean is not affected by the existence of an outlier. The median is not affected by the e

xistence of an outlier. The standard deviation is affected by the existence of an outlier. The interquartile range is unaffected by the existence of an outlier.
Mathematics
2 answers:
taurus [48]3 years ago
5 0
The correct answer, or incorrect statement, is A.  The mean is not affected by the existence of an outlier.  
choli [55]3 years ago
5 0

Answer:

The mean is not affected by the existence of an outlier.

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At one gym, there is a $12 start-up fee, and after that each month, m, at the gym costs $20. At another gym, each month at the g
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Answer: it will take 6 months for both gyms to cost the same.

Step-by-step explanation:

Let m represent the number of months that it will take for the cost of both gyms be the same.

At one gym, there is a $12 start-up fee, and after that each month, m, at the gym costs $20. This means that the total cost for m months is

12 + 20m

At another gym, each month at the gym costs $22. This means that the total cost for m months is 22m

For both costs to be the same, the number of months would be

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22m - 20m = 12

2m = 12

m = 12/2 = 6

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4 years ago
My friend and I have different calling plans. Mine has a monthly fee of $10 and $0.08 per minute charge. My friend pays $0.06 a
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difference is monthly customer fee= 15-10 = $5

difference in minutely charge = 0.08-0.06 = $0.02

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7 0
4 years ago
Two students have examined the scatter plot shown and have created a line of best fit for the data. Student A believes that the
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3 years ago
What is a reasonable estimate for the addition equation 1.5 + 0.9 = ________ when you round each addend to the nearest whole num
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The answer is C
hope this helps
8 0
3 years ago
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The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 8 cm/s. When the length is
iVinArrow [24]

Answer:

Step-by-step explanation:

A = lw \nl\frac{\dA}{\dt} =\frac{\dA}{\text{d}l}\frac{\text{d}l}{\dt} + \frac{\dA}{\text{d}w}\frac{\text{d}w}{\dt} =w\frac{\text{d}l}{\dt} + l\frac{\text{d}w}{\dt} \nll = 20 \text{ cm} \;\; \frac{\text{d}l}{\dt} = 8 \text{ cm/s} \;\;w = 10 \text{ cm}, \;\;  ]\frac{\text{d}w}{\dt} = 3 \text{ cm/s} \nl\frac{\dA}{\dt} =(  10 \text{ cm} )( 8 \text{ cm/s} ) + (  20 \text{ cm} )( 3 \text{ cm/s} )  =140 \text{ cm}^2\!\text{/s}


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3 years ago
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