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3 years ago

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On the first Saturday of every month, a local amusement park allows children under the age of 12 to ride for free. Which of thes

e children can ride for free: Jamie (12 years), Su (11 years, 9 months), Katalin (12 years, 3 months), Devon (13 years, 2 months)?

1 answer:

DochEvi [55]3 years ago

5 0

Su is the only child under 12 years of age, therefore, she is only child entering for free.

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Jonesville and smithville each have a population of size 2600 at time t = 0, where t is measured in years. Suppose jonesville'

ehidna [41]

Answer: B. Jonesville is growing linearly and Smithville is growing exponentially.

Step-by-step explanation:

Linear growth :

Population grow by a constant amount after each time period.
The rate of change of dependent variable with respect to independent variable is a constant.
It is represented by line on graph.
Equation for linear growth : $y=mx+c$ , c = initial value and m is the rate of change of y with respect to x.

Exponential growth :

Population grow by a constant ratio .
It is represented by a curve on graph.
Equation for exponential growth : $y=a(1+r)^x$ , a = initial value and r is rate of growth ( in decimal ) and x is time period.

Given : Jonesville's population grows by 170 people per year.

i.e .Population grow by a constant amount per year.

⇒ Jonesville is growing linearly.

The population of smithville grows by 7% per year.

i.e. Population grow by a constant ratio.

⇒Smithville is growing exponentially.

Hence, the true statement is "B. Jonesville is growing linearly and Smithville is growing exponentially."

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3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

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50%. Each coin has a 50% chance on landing on head and a 50% chance of landont on tails.

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Answer:

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Step-by-step explanation:

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