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erma4kov [3.2K]
2 years ago
5

For her phone service, Melissa pays a monthly fee of $16, and she pays an additional $0.06 per minute of use. The least she has

been charged in a month is $93.22. What are the possible numbers of minutes she has used her phone in a month? Use m for the number of minutes, and solve your inequality for m.​
Mathematics
1 answer:
Annette [7]2 years ago
8 0

Answer:

1287

Step-by-step explanation:

93.22-16=77.22

77.22 devided by 0.06=1287

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Cheryl collected data for her mathematics project. She noted that the data set was approximately normal.
nadya68 [22]

Answer:

X_(r) >> X_(n)

The mean for this case would increase since is defined as:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

The interquartile range would not change since the definition for the IQR is IQR =Q_3 -Q_1 and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

Step-by-step explanation:

For this case we assume that we have a random sample given X_(1), X_(2) ,..., X_(n) and for each observation X_i \sim N(\mu, \sigma) since the problem states that the data is approximately normal.

Let's assume that the largest value on this sample is X_(n) and for this case we are going to replace this value by another one extremely higher so we satisfy this condition:

X_(r) >> X_(n)

The mean for this case would increase since is defined as:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

The interquartile range would not change since the definition for the IQR is IQR =Q_3 -Q_1 and the quartiles are the same.

The standard deviation would not remain the same since by definition is:

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And since we change the largest value the deviation would increase considerably.

And for the last option is not always true since if we select a value so much higher then the distribution would be skewed to the right.

So the best option for this case is:

Mean would increase.

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3 years ago
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