Can the mean (average) be greater than the largest number in a data set?

1 answer:

3 0

No it can never be greater than the largest number in the set because an average (or a mean) is the "<span> number expressing the central or typical value in a set of data, in particular the mode, median, or (most commonly) the mean, which is calculated by dividing the sum of the values in the set by their number." c:</span>

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Lana draws a triangle( LMN) on the coordinate plane. What is the perimeter of the triangle( LMN)? Round to the nearest unit.

Dmitriy789 [7]

Answer:

Step-by-step explanation:

The longest side is 12.

The two other sides of the triangle= $\sqrt{3 squared+6 squared}$ =(9+36)square rooted=45 square rooted is about 6.7.

6.7*2=13.4

13.4+12=25.4

25.4 rounded to the nearest unit is 25.

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3 years ago

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disa [49]

Answer:

B. The amount spent on grapes compared with the weight of the purchase.

Step-by-step explanation:

Grapes are usually sold at some dollar amount per pound. That dollar amount is the "rate of change," and it is generally constant.

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In the case of pizza delivery or bus ridership, it is hard to imagine what the "rate of change" might be, as the relationship is probably <u>not</u> a function.

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3 years ago

Upgrading a certain software package requires installation of 68 new files. Files are installed consecutively. The installation

ser-zykov [4K]

Answer:

The probability that the whole package is uppgraded in less then 12 minutes is 0,1271

Step-by-step explanation:

The mean distribution for the length of the installation (in seconds) of the programs will be denoted by X. Using the Central Limit Theorem, we can assume that X is normal (it will be pretty close). The mean of X is 15 and the variance is 15, hence, the standard deviation is √15 = 3.873.

We want to find the probability that the full installation process takes less than 12 minutes = 720 seconds. Then, in average, each program should take less than 720/68 = 10.5882 seconds to install. Hence, we want to find the probability of X being less than 10.5882. For that, we will take W, the standariation of X, given by the following formula

$W = \frac{X-\mu}{\sigma} = \frac{X-15}{3.873}$

We will work with $\phi$ , the cummulative distribution function of the standard Normal variable W. The values of $\phi$ can be found in the attached file.