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how do i calculate the population growth while using the log function and doubling time

Double 5711767=(1.004)n 5711767(1+.004)n 11423534=5711767(1.004) 2=(1.004)n from here I get lost

Research, Knowledge and Information :

Doubling time - Wikipedia

The doubling time is the period of time required for a quantity to double in size or value. It is applied to population growth, inflation, resource extraction ...
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How to properly do a population doubling ... - ResearchGate

How to properly do a population doubling? ... the growth curve follows a Gompertzian function rather ... how does one calculate doubling time of a cell line using ...
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How to Calculate Growth Rate or Percent Change | Sciencing

How to Calculate Growth Rate or Percent Change ... differences due to change over time, such as population growth. ... to Calculate the Time for Cell Doubling;
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Calculating Rate and Exponential Growth: The Population ...

To find the population growth as a function of time, ... I get the natural log ... Calculating Rate and Exponential Growth: The Population Dynamics Problem Related ...
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Exponential growth - Wikipedia

Exponential growth is ... it is also called geometric growth or geometric decay, the function ... This means that the doubling time of the American population ...
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Doubling Time - Continuous Compounding - finance formulas

The doubling time formula with continuous compounding is the natural log of 2 ... Using the doubling time for ... Formulas related to Doubling Time
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Initial Value Problems for Growth and Decay

Initial Value Problems for Growth ... the exponential function, we take the natural log of ... the doubling time directly, but we can calculate it from the ...
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ATCC ANIMAL CELL CULTURE GUIDE

ATCC® ANIMAL CELL CULTURE GUIDE ... Lag, log or exponential, stationary ... Calculate the population doubling time, ...
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Suggested Questions And Answer :

Use a growth rate of 7% to predict the population in 2074 of a country that in the year 2006 had a population of 100 million?

2074-2006=68 years. If the growth rate is 7% per year then the growth factor is 1.07^68=99.56275 approx. Multiply by 100,000,000=9,956,275,000. (1+r)^2=1+2r+r^2=1+2r approx for small r. This is the doubling formula. 68=64+4=2^6+2^2. If we apply the formula 6 times we get an approximate value for 1.07^64 (r=0.07): 1.07^2=1+2*0.07=1+0.14=1.14 (true value: 1.1449) Now put r=0.14 1.07^4=1.14^2=1.28 (1.3107...) 1.07^8=1.28^2=1.56 (1.7181...) 1.07^16=1.56^2=1+1.12=2.12 (2.9521...) 2.12=2*1.06=2(1+0.06); 1.07^32=2^2*1.06^2=4*1.12=4.48 (8.7152...) 4.48=4*1.12; 1.07^64=4^2(1.12)^2=16*1.24=16+3.84=19.84 (75.9559...) 1.07^68=1.07^64 * 1.07^4=19.84*1.28=25.3952 (99.56274...) So, using the doubling formula we have an underestimate of 100,000,000*25.4=2,540,000,000 instead of about 10,000,000,000. The doubling time formula should give a better approximation: when the growth is double (1+r)^T=2 where T is the time for growth to be doubled, so T=log(2)/log(1+0.07)=ln(2)/0.07 approx.=9.9 years. Call this 10 years. In 68 years then, the population should have doubled about 6.8 times. Let's work out doubling 6 times: 2^6=64 so after 6*10=60 years (2066) the population will be 6,400,000,000. Now, using our approximate doubling formula from earlier we can estimate for the remaining 8 years. We simply multiply by 1.56 (1.07^8) to give 9,984,000,000, which is very close to the true value of 1.07^68 * 100,000,000. If we use simple proportion to find the growth over the final 8 years, we would use the factor 8/10 * 2=1.6. This produces 1.6*64,000,000,000=102,400,000,000, which is an overestimate.
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how do i calculate the population growth while using the log function and doubling time

find out how many dubels yu hav tipikal problem =gro av a lab kulture....a mold or bakteria will dubel in time=such & such So how big will it be tumoro morn? If it dubel evree half-our & yu kum tu werk 9 ours later... now/start=2^18 See also...radioaktive dekae hav a time tu kut in half
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Use different P and t, plot delta P/delta t against P n obtain the approximate value for r n k.

Inspection of the derivative function shows that it's a parabola​. When P=0 or k, dP/dt=0. And we can rewrite the function: P'=dP/dt=(rP/k)(k-P)=(r/k)(Pk-P^2)=(r/k)(k^2/4-P^2+Pk-k^2/4)=(r/k)((k/2)^2-(k-P)^2). The vertex is at (k/2,rk/2). So the distance between the point where the curve cuts the P (horizontal) axis gives us a measure for k; and, using this value for k we get r by inspecting the vertex height above the P axis and dividing by k/2. The table shows P against t, but we need P' against P; but we can roughly calculate dt and dP by subtracting pairs of numbers. When P=800 in the table we clearly have two values for t and the gradient P'=0. Therefore, k=800, implying that k/2=400. When P=400 in the table we can see that t is between 140 and 182 and P is between 371 and  534. So the gradient is roughly (534-371)/(182-140)=163/42=3.9. Therefore rk/2=3.9 so r=3.9/400=0.01. dP/P(1-P/k)=rdt. So int(dP/P(1-P/k))=rt+C, where C is to be determined from the table. Int(dP/P)-int(dP/(P-k))=rt+C; ln(P)-ln|P-k|=rt+C; ln(P/|P-k|)=rt+C; P/|P-k|=ae^rt where a replaces constant C (a=e^C). P=|P-k|ae^rt, so P=ake^rt/(1+ae^rt). From the table, when t=0, P=49, so 49=ak/(1+a); 49+49a=ak and a(k-49)=49, a=49/(k-49). Using the approximate value of k=800, a is approximately 49/(800-49)=0.065. dP/dt=rP-rP^2/k. Since r and k are constants the DE can be solved by separation of variables: So putting in our approximate values P=0.065*800*e^0.01t(1+0.065e^0.01t)⇒ P=52e^0.01t(1+0.065e^0.01t). The values in the table can be plotted on a graph of P against t, but to plot dP/dt against P would require us to know r and k, for which we have only the approximate values 0.01 and 800.
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(please show how you got answer)(please read the more information box)

dp/dt=50t^2-100t^3/2, so dp=(50t^2-100t^3/2)dt. Integrating we get p=(50t^3)/3-(200/5)t^5/2+p0, where p0 is an initial population. We have no indication of p0, so we have to assume it's zero or negligible at t=0. So p=(50t^3)/3-40t^(5/2)=50000. Using a calculator, t=18.856 years for the population to reach 50000. Using a graph it is clear that at t=0, p=0, and between p=0 and 5.76 the population is negative. After t=5.76 years the population rises steadily until at t=18.86 it reaches 50000. At t=4 the graph appears to have a minimum at p=-213.33. To offset this p0 could be set to 213.33 so that the minimum is zero at t=4 years. When this adjustment is made the population reaches 50000 at t=18.836 years approximately.
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discuss by considering the sign of dP/dt ,the relationship between P n k if P0 < k and if P0 > k.

dP/dt<0 when P/k>1, P>k and dP/dt>0 when Pk it decreases.
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homework help

17 ft/sekond is SPEED, not velosity
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poisson distribution

PDF: f(X)=(1/mu)e^-(X/mu); lambda=1/mu where mu is the mean time between events, so lambda is the number of events in unit time. X>0. f(X)=0 when X=0. CDF: F(X)=1-e^-(X/mu), where mu=E(X) (expectation) a) mu=10 mins=1/3 of a half-hour. Lambda=3. f(1)=3e^-3=0.1494 or 14.94% b) f(X)=0 when X=0 by definition, so we cannot use f(X) directly, so we use lambda=6, the number of calls within an hour, and then we can calculate the probability using the CDF function F(X): P(X>1/2)=1-P(X<=1/2)=1-F(1/2)=1-(1-e^-(6/2))=e^-3=0.0498 approx, probability of no calls within half an hour. c) If probability is 0.02 of no calls in x hours we use the same logic as in b): Again we start with P(X>x)=1-P(X<=x)=1-F(x)=1-(1-e^-(6x))=e^-(6x)=0.02, so, taking natural logs: -6x=ln(0.02)=-3.912, so x=0.652hr=39.12 mins approx. d) We know from (b) the probability of no calls within a half-hour is e^-3. The same probability applies to each of the non-overlapping half-hour periods, so the combined probability is (e^-3)^4=e^-12=0.000006144 approx. e) There is a factor of 3 between (a) and (b): (a)=3(b). More...
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Solve the following without using calculator

You need log tables or a calculator for an accurate answer. log(2)+log(5)=1, so if log(2)=0.3 approx then log5 is 1-0.3=0.7 approx. (We know this because 2^10=1024 which is approximately 10^3. So 2=10^(3/10)=10^0.3, making 0.3=log[10]2 approximately, using [] to show the base.) However, we don't know the log to base 10 of 23 or 45 without tables or a calculator. 23 lies between 16 (2^4) and 32 (2^5) so its log to base 2 is between 4 and 5. 45 is between 25 (5^2) and 125 (5^3), so its log to base 5 is between 2 and 3. Therefore the result is roughly 4.5+2.5=7. Another way of estimating the log is to use a linear correlation. Take y=log[5](x). When x=5 y=1, and when x=25 y=2, and when x=125, y=3. Consider the number 45. It lies between 25 and 125. Consider a right-angled triangle with height 1 and base 100. The base represents the gap between 25 and 125 and the height represents the difference of 1 between the logs of 25 and 125 (2 and 3 respectively). The slope of the triangle is 1/100=0.01. The difference between 45 and 25 is 20, so this is one fifth the way between 25 and 125. Therefore the log is about a fifth of the distance between 2 and 3, making it about 2.2 (1/5=0.2). So log[5](45) is about 2.2. Now consider y=log[2](x). 23 is between 16 and 32 so y is between 4 and 5. This time the slope of the triangle is 1/16. 23-16=7, so 23 is 7/16 along the triangle's base. 7/16 of the height is about 0.4, so we add that to 4 to get 4.4 and we have the approximate log[2](23). Add this to 2.2 and we get 6.6. This is another approximation to the answer without using a calculator or tables. [Using a calculator the answer is about 6.8888. 6.6 is accurate to about 96%, which isn't bad. Our first guesstimate of 7 was over by less than 2%. An average of the two guesses gives us about 99% accuracy.]
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Explain the results of the following options: Option 1: 6% compound interest quarterly for 5 years. Option 2: 8% compound interest annually for 5 years. Option 3: 14.5% simple interest for 10 years.

MEMO TO CLIENT Simple interest applies the interest rate proportionately, so the amount of interest on a particular investment is directly proportional to the length of time invested. This means that, for example, if the investment period is 5 years, the interest is 5 times the interest earned in one year; for 10 years it is 10 times that earned in a year. It is also easy to calculate because of this simple proportion. Compound interest is more rewarding to the investor. After a period of time, for example, a year, the interest earned in the year is added to the original amount invested. So at this point, it is the same as simple interest. But what happens next is different. The investment plus the interest becomes the invested amount for the next period, the next year, for example. At the end of this period the process continues, and the interest is again added and becomes the investment amount for the next period. So it is clear that over a period of time more and more interest is earned. An important feature that investors need to be aware of is: how regularly is compound interest added? The shorter the period, the bigger the interest earned. Interest can be compounded annually, quarterly, monthly, daily or continuously. So, if the investor is quoted a particular annual rate of interest, then the largest amount of interest gained will correspond to the shortest compound interest period. As an example, take 6% per annum, or annual interest rate. After a year with interest compounded annually, 6% interest will be earned. If interest is compounded quarterly, then each quarter the interest will be added at a rate of 1.5% each quarter, but by the end of a year, the effective interest will be more than 6.1%. If interest is compounded monthly, the monthly rate would be 0.5% and after a year would be effectively closer to 6.2%. Interest compounded daily would be even closer to 6.2% and continuously would be slightly more. Growth is a convenient way of expressing the factor by which an investment increases over time, and companies will often publish tables to simplify calculations of expected returns on investments at fixed rates. The time periods will be typically 5, 10, 15, 25 years for a range of annual rates. So investors can quickly calculate the returns on varying amounts of money. As an example, take 15 years. The growth rates at 6% per annum would be: 1.9 (simple interest); 2.40 (compounded annually); 2.44 (compounded quarterly); 2.45 (compounded monthly); 2.46 (compounded daily or continuously). Option 1 6% annually is 1.5% quarterly, so growth is 1.015^20=1.3469, where 20 is 20*(1/4)=5 years. \$500000*1.3469=\$673,427.50 to best accuracy. Option 2 8% compounded annually: growth=1.08^5=1.4693 and amount is \$500000*1.4693=\$734,664.04. Option 3 14.5% simple interest for 10 years: 145% interest=1.45*500000=\$725,000 interest+500000=\$1.225 million. The first two options have the same investment time period, and option 2 is better. Option 3 has double the time period. If option 3 were to be applied over 5 years instead of 10, the interest would have been \$362,500 (half of \$725,000) and the total amount would have been \$862,500. However, the investment is over ten years so the investor would need to wait 10 years before taking full advantage of the investment. Take option 2 over 10 years and we get a growth rate of 2.1589 making the investment worth \$1,079,462.50, which is still smaller than option 3, which had a growth rate of 2.45 (1.45+1) because of the higher interest rate.
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What's a function of function?

If you just mean the purpose of a function then see immediately below. If you mean f of g where f and g are two functions, go to the end. The purpose of a function is like giving someone a to-do list. A function (normally shown as f(x)= meaning "function of x", or y=) will usually contain one variable, usually represented by x. Some functions may have more than one variable. If x isn't the variable, it will usually be a different letter like t or a or any letter. Let's say it is x. This variable will appear in a formula with numbers. The formula is the function and it contains instructions in symbols telling you what to do with the variable. For example, multiply the variable by 2 then add 3 and divide the result by 4. This would be written f(x)=(2x+3)/4 or y=(2x+3)/4. The equals sign means that the function is defined as the expression on the right. Just like someone might say to you, think of a number (that's x), but don't tell me what it is, then double it and add 3 and divide the result by 4. Functions can be plotted on a graph. The idea of this is that on the horizontal axis (x axis) you have markers for 0, 1, 2, etc. On the vertical axis (y or f(x)) you also have markers. The graph is usually a continuous line, and every point on the line is made by putting a different value for x and marking the point (on a rectangular grid f(x) units vertically and x units horizontally) for the result of working out the value of the function for different values of x. The points make up the graph. You can use the graph to find the value of x for any value of the function, and the value of the function for any value of x, provided the graph extends far enough. This is just a brief example of the purpose of a function. A practical example is the conversion of temperature in degrees Fahrenheit (F) to degrees Celsius: C=5(F-32)/9. This function tells you what do with the value of the temperature. You could plot this as a straight line graph and you can read off degrees C for any temperature in degrees F, and vice versa. Another example is d=30t where d=distance, speed=30mph, t=time. The function is 30t, and from it you can work out the distance a car moves in a particular time t when its speed is 30mph. Function of a function: this simply means you work out the value of applying one function and then feed this answer as the variable into the other function. An example could be that one function is used to find out how many miles a car travelling at an average speed of 30mph in a given time. The second function could be to find out how much fuel is used to travel that distance. So g(t)=30t is the first function. The second function is h(d)=d/24 where fuel consumption is 24 miles per gallon. h of g is h(g(t))=30t/24=5t/4.
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