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kakasveta [241]
3 years ago
11

Given an initial quantity Q0=150 and a growth rate of 7% per unit time, give a formula for the quantity Q as a function of time

t, and find the value of the quantity at t=10.
a) Assume the growth rate is not continuous. Round your answer for Q(10) to three decimal places.
Q(t)=
Q(10) =
b) Assume the growth rate is continuous. Round your answer for Q(10) to three decimal places.
Q(t)=
Q(10)=
Mathematics
1 answer:
rodikova [14]3 years ago
8 0

Answer:

Step-by-step explanation:

From the given information:

(a)

Since growth quantity is not continuous

Q(t) = 150 (1.07)^t

For t = 10

Q(10) = 150 (1.07)^{10}

\mathbf{Q(10) = 295.073}

(b)

Here, for a continuous growth rate, the growth quantity can be computed in terms of initial quantity and the growth rate.

i.e.

Q(t) = 150 e^{0.07t}

At t = 10 for a continuous growth rate;

Q(10) = 150 e^{0.07  \times 10}

Q(10) = 150 e^{0.7}

\mathbf{Q(10) = 302.063}

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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

x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13

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}

\bar x = 12.64

Next, the sample variance is:

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

So, we have:

\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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