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3 years ago

Explain why the initial value of any function of the form f(x) = a(bx) is equal to a.

1 answer:

Montano1993 [528]3 years ago

7 0

So the said problem asking to explain or justify why the initial value of any function of the form f(x) = a(bx) is equal to a. For me I won't agree with the said question, the said function is determine through the value of X because it is the undefined value in the equation that is needed to be define. I hope this would help

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Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

8 0

3 years ago

Read 2 more answers

Solve 3x^2 + 6x+15=0

Elden [556K]

Answer:

The solution of given equation is ( - 1 + 2 i ) , ( - 1 - 2 i )

Step-by-step explanation:

Given equation as :

3 x² + 6 x +15 = 0

The value of x fro the quadratic equation a x² + b x + c = 0 is obtained as

x = $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

So , from given eq , the value of x is now obtain as

x = $\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

Or, x = $\frac{-6\pm \sqrt{6^{2}-4\times 3\times 15}}{2\times 3}$

Or, x = $\frac{\sqrt{-144} }{6}$

∴ x = ( - 1 + 2 i ) , ( - 1 - 2 i )

Hence The solution of given equation is ( - 1 + 2 i ) , ( - 1 - 2 i ) Answer

6 0

3 years ago

Determine if each graph represent a direct proportional or not

Troyanec [42]

Answer:

i hope everyone has a good day keep your head up at all times and dont give up

Step-by-step explanation:

5 0

3 years ago

Use the graph below to write the equation of the function

Ksivusya [100]

Answer:

Step-by-step explanation:

gefhhfeyjhtwgi15287

8 0

3 years ago

According to a recent publication, the mean price of new mobile homes is $63 comma 800. Assume a standard deviation of $7900.

wolverine [178]

Answer:

a. For n=25, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are $63,800 and $1,580, respectively.

b. For n=50, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are $63,800 and $1,117, respectively.

Step-by-step explanation:

In this case, for each sample size, we have a sampling distribution (a distribution for the population of sample means), with the following parameters:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{n}$

For n=25 we have:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{25}=7,900/5=1,580$

The spread of the sampling distribution is always smaller than the population spread of the individuals. The spread is smaller as the sample size increase.

This has the implication that is expected to have more precision in the estimation of the population mean when we use bigger samples than smaller ones.

If n=50, we have:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{50}=7,900/7.07=1,117$

4 0

4 years ago