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4 years ago

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Which choice could not be a set of intervals for a frequency table of histogram?

1 answer:

valentina_108 [34]4 years ago

6 0

When constructing the histogram, the intervals are selected in such a way that the width of each interval is the same and the gap between two intervals stays uniform throughout the histogram. This makes the data easy to visualize, easy to interpret and easy to understand.

From the given options, option B lists the intervals in an absurd order. First interval is from 0 to 3. There is no gap between first and second interval and the second interval is larger than the first. So these intervals cannot be the set of intervals of a histogram.

So, the answer to this question is option B

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The population P of a small town, measured in hundreds, is modeled by the inverse of the function P−1(t) = 10ln(50t − 1000), whe

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Using the inverse function, it is found that the equation is:

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---------------------------

We want to find the inverse of:

$y = P^{-1}(t) = 10\ln{50t - 1000}$

To do this, we <u>exchange y and t, and isolate y.</u>

Then:

$10\ln{50y - 1000} = t$

$\ln{50y - 1000} = \frac{t}{10}$

$e^{\ln{50y - 1000}} = e^{0.1t}$

$50y - 1000 = e^{0.1t}$

$50y = e^{0.1t} + 1000$

$y = \frac{e^{0.1t} + 1000}{50}$

$y = \frac{e^{0.1t}}{50} + \frac{1000}{50}$

$P(t) = \frac{e^{0.1t}}{50} + 20$

It's initial value is:

$P(0) = \frac{e^{0.1(0)}}{50} + 20 = 0.02 + 20 = 20.02$

Rounding, 20.

A similar problem is given at brainly.com/question/23950969

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