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aalyn [17]
2 years ago
8

The distance from Michaels house to his Granma house is 84 miles round trip . If Michael visits his granma 9 times each year how

many miles does he travel to and from his grandmas each year.
Mathematics
2 answers:
kvv77 [185]2 years ago
4 0

Answer:

Michael travels a total of 756 miles each year.

Step-by-step explanation:

If you want the total number of miles he travels(to and from the house combined), just multiply 84 by 9.

84 x 9 = 756

prohojiy [21]2 years ago
4 0

Answer:

Step-by-step explanation:

Michael travels a total of 756 miles each year.

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kati45 [8]

Answer:

Neither the ranges nor the interquartile ranges for the data sets are the same.

Step-by-step explanation:

In a visual display, the boxplot presents five sample statistics: the minimum, the lower quartile, the median, the upper quartile and the maximum, and the box length gives an indication of the sample variability and the line across the box shows where the sample is centred, with an end at each quartile. The length of the box is thus the interquartile range of the sample and, whether the sample is symmetric or skewed, either to the right or left, the "shape" of the sample, and by implication, the shape of the population from which it was drawn, considering appropriate analyses of the data.

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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:

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3 years ago
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Sedbober [7]

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The answer is: \frac{(x-3)(x+1)}{2x^{2}(x+3)(x+2)}

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Answer:

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Step-by-step explanation:

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