The table represents the function f(x).

Now, you will consider Option 1, setting a maximum shower time of 10 minutes.

matrenka [14]

The maximum shower time is an illustration of mean and median, and the conclusion is to disagree with Blake's claim

<h3>How to interpret the shower time?</h3>

The question is incomplete, as the dataset (and the data elements) are not given.

So, I will answer this question using the following (assumed) dataset:

Shower time (in minutes): 6, 7, 7, 8, 8, 9, 9, 9, 12, 12, 12, 13, 15,

Calculate the mean:

Mean = Sum/Count

So, we have:

Mean = (6+ 7+ 7+ 8+ 8+ 9+ 9+ 9+ 12+ 12+ 12+ 13+ 15)/13

Mean = 9.8

The median is the middle element.

So, we have:

Median = 9

From the question, we have the following assumptions:

The shower time of students whose shower times are above 10 minutes, is 10 minutes
Other shower time remains unchanged.

So, the dataset becomes: 6, 7, 7, 8, 8, 9, 9, 9, 10, 10, 10, 10, 10

The mean is:

Mean = (6+ 7+ 7+ 8+ 8+ 9+ 9+ 9+ 10+ 10+ 10+ 10+ 10)/13

Mean = 8.7

The median is the middle element.

So, we have:

Median = 9

From the above computation, we have the following table:

Initial Final

Mean 9.8 8.7

Median 9 9

Notice that the mean value changed, but it did not go below 8 as claimed by Blake; while the median remains unchanged.

Hence, the conclusion is to disagree with Blake's claim

Read more about mean and median at:

brainly.com/question/14532771

#SPJ1

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B , the second option, is your answer

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How to write an equivalent expression for n time a using o

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It would have to be B

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Read 2 more answers

How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy

lisov135 [29]

Answer:

$300\; \rm lb$ .

Step-by-step explanation:

Let $x$ represent the mass (in pounds) of that $15\%$ copper alloy required, such that the final mixture would contain $25.5\%$ copper by mass.

Consider: if $x$ pounds of that $15\%$ copper alloy is mixed with $700$ pounds that $30\%$ copper alloy, what would be the mass of copper in the mixture?

Mass of copper in $x$ pounds of that $15\%$ copper alloy: $(0.15\, x)\; \rm lb$ .
Mass of copper in $700$ pounds of that $30\%$ copper alloy: $700 \times 0.30 = 210\; \rm lb$ .