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2 years ago

Can all linear functions be modeled by an arithmetic sequence? if not, provide a counterexample

1 answer:

4 0

It depends. Generally no.

Linear equations are generally in the form [math]y=mx+b[/math] and have a domain of [math](-\infty,\infty)[/math], or all real numbers. However, an arithmetic sequence is only defines for the natural numbers (that is, while numbers [math]> 0[/math].

For any two terms in the arithmetic sequence, [math]a_n[/math] and [math]a_{n+1}[/math], there will always be a point on the linear function that lies in between them, and is such not defined in the sequence.

This does not make the sequence and function unrelated, but rather it makes them not the same.

A similar argument applies for geometric sequences and exponential equations.

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You put $1200 in an account that earns 3% simple interest. Y= a (1 + r)t Find the total amount in the account after four years.

olchik [2.2K]

Hello!

To find how much money total will be in the account after four years, you need to use the function, $y = a(1 + r)^{t}$ , as seen above. In this function, a is the starting value, r is the interest rate, and t is the amount of time (in years).

Before substituting values into the function, we need to convert 3% into a decimal. Remember that all percentages are out of 100, so we divide 3 by 100 and remove the percentage sign, or move the decimal two places to the left.

3/100 = 0.03 | 0.03 is the interest rate.

With all the necessary values, we can find the total amount in the account. In this case, a = 1200, r = 0.03, and t = 4.

$y = 1200(1 + 0.03)^{4}$

$y = 1200(1.03)^{4}$

$y = 1200(1.12550881)$

y = 1350.610572, which can be rounded to 1350.61 dollars.

Therefore, after 4 years, the total amount in the account will be about 1350.61 dollars.

6 0

2 years ago

Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take

MrRissso [65]

Answer:

mean (μ) = 4.25

Step-by-step explanation:

Let p = probability of a defective computer components = $\frac{68}{400} = 0.17$

let q = probability of a non-defective computer components = $\frac{332}{400} = 0.83$

Given random sample n = 25

we will find mean value in binomial distribution

The mean of binomial distribution = np

here 'n' is sample size and 'p' is defective components

mean (μ) = 25 X 0.17 = 4.25

<u>Conclusion</u>:-

mean (μ) = 4.25

6 0

2 years ago

Karina read a total of 20 2/4 pages in her science and social studies books combined she read 12 3/4 pages in her science book h

wlad13 [49]

Answer:

8 1/4

Step-by-step explanation:

20 2/4

12 3/4

__________

8 1/4

4 0

2 years ago

Read 2 more answers

A tennis coach divides her 9-player squad into three 3-player groups with each player in only one group. How many different sets

Lostsunrise [7]

Answer:

Step-by-step explanation:

The coach divides her 9-player squad into 3-player groups. This means that she has 9 players and she wants to share them into 3s.

The number sets of three groups she will have can be obtained by dividing the total number of her players by the 3. That is:

9 / 3 = 3 sets

Therefore, she will have 3 sets of 3-player groups.

7 0

2 years ago

Investigate the following harvesting model both qualitatively and analytically. If a constant number h of fish are harvested fro

zalisa [80]

Answer:

a. The population does not become extinct in finite time.

Step-by-step explanation:

The model for the population of the fishery is

$dP/dt = P(a-bP)-h, P(0) = P_0$

If we rearrange and replace the constants we have:

$\frac{dP}{P(7-P)-49/4} =dt\\\\-4 (\frac{dP}{4(P-7)P+49}) =dt\\\\-4 \frac{dP}{(2P-7)^2} =dt\\\\-4 \int\frac{dP}{(2P-7)^2} =\int dt\\\\-4(-\frac{1}{2(2P-7)})=t+C\\\\\frac{2}{2P-7}=t+C\\\\ t=0 \,\,\, P(0)=P_0\\\\\frac{2}{2P_0-7}=0+C\\\\C=\frac{2}{2P_0-7}$

Now we can calculate if the population become 0 in any finite time

$\frac{2}{2P-7}=t+\frac{2}{2P_0-7}\\\\\frac{2}{2*0-7}=t+\frac{2}{2P_0-7}\\\\-\frac{2}{7}=t+\frac{2}{2P_0-7}\\\\$

To be a finite time, t>0

$t=-\frac{2}{7}-\frac{2}{2P_0-7}=0\\\\-\frac{2}{2P_0-7}=\frac{2}{7}\\\\7-2P_0=7\\\\P_0=0$

We can conclude that the only finite time in which P=0 is when the initial population is 0.

Because P0 is a positive constant, we can say that the population does not become extint in finite time.

5 0

2 years ago