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3 years ago

a population is growing continuously at a rate of 5%. if the population is now 3400, what will it be in 17 years?

Mathematics

2 answers:

DedPeter [7]3 years ago

6 0

Answer:

7793

Step-by-step explanation:

3400 x 1.05^17 = 7792.862

Katena32 [7]3 years ago

4 0

Answer:

The population would be 7792.86 or when rounded would be 7793

Step-by-step explanation:

This question is an example of a repeated percentage change

To work this out you would first have to convert the percentage of 5 into a decimal. You can do this by dividing 5 by 100, this gives you 0.05. This is because percentages are out of 100.

The next step is to add 0.05 to 1, which gives you 1.05. This is because we are increasing the amount by 5% annually.

The final step is to multiply the amount of 3400 by 1.05 to the power of 17, this gives you 7792.86. This is because we are finding 5% and adding it to the amount and finding 5% of that new amount, this will continue 17 times.

1) Divide 5 by 100.

$5/100=0.05$

2) Add 0.05 to 1.

$0.05+1=1.05$

3) Multiply 3400 by 1.05 to the power of 17.

$3400*1.05^{17} = 7792.86$

4) Round to the nearest whole number.

7793

You might be interested in

Which fraction is equal to 5/6? 8/22, 25/30, 5/15?

nadya68 [22]

If the fraction cannot be simplified any more, you can multiply both the numerator and denominator by an number.

5/6 (multiply the top and bottom by 2) is the same as 10/12
8/22 (can be simplified by dividing top and bottom by 2) is 4/11
25/30 (simplified by dividing by 5) is 5/6
5/15 (simplified by dividing by 5) is 1/3

Hope this helps!

3 0

3 years ago

Show that if X is a geometric random variable with parameter p, then

Lubov Fominskaja [6]

Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: