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ArbitrLikvidat [17]
3 years ago
6

A population grows exponentially according to the differential equation dP, dt equals k times P , where P is the population, t i

s time, and k is a positive constant. If P(0) = A, what is the time for the population to quadruple its initial value?
Mathematics
1 answer:
guajiro [1.7K]3 years ago
5 0

Answer:

\frac{\ln 4}{k}

Step-by-step explanation:

Given equation that shows the change in population with respect to time,

\frac{dP}{dt}=kP

\frac{dP}{P}=kdt

On integrating,

\ln P = kt + C

\implies P = e^{kt+C}

P=e^C e^{kt}

Put e^C=P_0

P = P_0 e^{kt}

According to the question,

If P = A if t = 0,

A = P_0 e^0\implies A = P_0

Thus, the required function,

P = A e^{kt}

If the final population = 4A,

4A = A e^{kt}

4 = e^{kt}

Taking natural log both sides,

\ln 4 = \ln e^{kt}

\ln 4 = kt\ln e

\ln 4 = kt

\implies t = \frac{\ln 4}{k}

Hence, after \frac{\ln 4}{k} time, the population will be quadruple its initial value.

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Which of the following is a solution of x2 + 4x + 25?
blondinia [14]
Solve for x:
x^2 + 4 x + 25 = 0     I ssume that's the notation.
Subtract 25 from both sides:
x^2 + 4 x = -25
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x^2 + 4 x + 4 = -21
Write the left hand side as a square:
(x + 2)^2 = -21
Take the square root of both sides:
x + 2 = i sqrt(21) or x + 2 = -i sqrt(21)
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what are you answers ?

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1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
Ad libitum [116K]
Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

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

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

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

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

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

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21
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Answer:

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24 + 25 > 7

7 + 25 > 24 therefore it is indeed a right triangle

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