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Alex
3 years ago
13

Paul is gathering data about moss growth in a local forest. He measured an area of 11 square centimeters on one particular tree

and will come back in 6 months to measure the growth of the moss. If the area covered by moss multiplies by one and a half times each month, approximately how much area will the moss cover when Paul returns?
Mathematics
2 answers:
STatiana [176]3 years ago
8 0
If you so 11 * 1.5 * 1.5 * 1.5 * 1.5 * 1.5 * 1.5 you get approximately 125.2969cm squared
Nikitich [7]3 years ago
4 0
<h2>Answer:</h2>

The area that moss cover when Paul returns is:

              125.296875\ \text{square\ centimeters}

<h2>Step-by-step explanation:</h2>

The initial area that is measured by Paul about moss growth in a local forest is:  11 square centimeters.

We know that the increase in the area covered by moss is increasing exponentially.

Since each month it is increasing by a fixed constant 1.5 times as it was the previous month.

Hence, the function which represents the area covered by moss in x months is:

     A(x)=11\times (1.5)^x

We are asked to find the value of function when x=6

i.e.

A(x)=11\times 1.5\times 1.5\times 1.5\times 1.5\times 1.5\times 1.5\\\\\\A(x)=125.296875\ \text{square\ centimeters}

            Hence, the answer is:

            125.296875\ \text{square\ centimeters}

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Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl
lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13

x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88

x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63

x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38

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So, the table becomes:

\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}

Next, calculate the mean

\bar x = \frac{\sum fx}{\sum f}

\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}

\bar x = \frac{1845.73}{146}

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Next, the sample variance is:

\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}

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\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}

\sigma^2 = \frac{2393.6875}{145}

\sigma^2 = 16.51

The sample standard deviation is:

\sigma = \sqrt{\sigma^2}

\sigma = \sqrt{16.51}

\sigma = 4.06

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