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3 years ago

How much money will you make if you make 11.35 for 10 hours 4 days a week?

1 answer:

4 0

Correct me if im wrong, if you multiply 11.35 by 10 you get $113.5. multiply by 4 and that equals to $454.

-hope this helps

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kotegsom [21]

Answer:

Step-by-step explanation:

i think it is 2

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3 years ago

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The ages of the winners of a cycling tournament are approximately bell-shaped. The mean age is 27.2 years, with a standard devia

elena-14-01-66 [18.8K]

Answer:

a. z-score = -1.2

b. The age 23 years old is -1.2 standard deviations below the mean

c. The age is not unusual

Step-by-step explanation:

a. Transformation to z-score

Mathematically;

z-score = (x - μ)/σ

where σ and μ are the standard deviation and the mean respectively.

From the question, x = 23, μ = 27.2 and σ = 3.5

Let’s put these into the equation;

z-score = (23-27.2)/3.5 = -1.2

b. Interpretation of result

The interpretation is that 23 years old is -1.2 standard deviations below the mean

c. Determination

The age is not unusual. This is because any z-value below -2 or above 2 is considered unusual

-1.2 is above -2 and thus it is not unusual

7 0

2 years ago

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

2 years ago

a botanist was measuring how tall her plant grew. after two weeks it had grown 9.62 inches. the second week alone it had grown 3

Elanso [62]

Answer:

6.5 inches

Step-by-step explanation:

Take the total 9.62 inches

Remove the 3.12 inches from the second week and there is your answer!

6 0

3 years ago

Sa piajes egne baseball games this season. Her point totais for each game were 8. 14.4 7.6.14, 4 and 7. What was

levacccp [35]

Answer:

The mean number of scores per game is 8

Step-by-step explanation:

Since we know we can find the mean by adding all the numbers and dividing them by many numbers there were. Once you add 8, 14, 4, 7, 6, 4, and 7, you will get 64. Since there were 8 numbers, you divide 64 by 8 to get 8.

3 0

3 years ago