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2 years ago

HELP PLEASE 20points!!!

Mathematics

2 answers:

labwork [276]2 years ago

7 0

Answer:

61%

Step-by-step explanation:

We can see that out of all the people that were surveyed, 54% were 10th graders. Since 33% out of all the ones surveyed were 10th graders that chose robotics, the fraction would be 33/54 which is 0.611.

This is 61% approx.

Furkat [3]2 years ago

5 0

Answer:

61%

Step-by-step explanation:

A P E X

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Which could be the entire interval over which the function,

Marrrta [24]

Answer:

(-2, 1)

Step-by-step explanation:

Hi Juliadayx! How are you?

Well, maybe you already heard about the domain and the range of a function.

The domain is the set of values that the independent variable can take (usually referred to as the letter "x"), while the range is the set of values that the dependent variable takes, which is called f(x) or function of x.

In this case, the exercise asks you to evaluate for which values of “x” (right column of the table), the function “f(x)” takes positive values (left column of the table), the positive values also include zero. And in this case you can see that the function f(x) is only positive for the values of "x": -2, -1, 0 and 1. Therefore, the answer is the entire interval (-2, 1).

I hope I've been helpful!

Regards!

3 0

3 years ago

Set up the integral that represents the arc length of the curve f(x) = ln(x) + 5 on [1, 3], and then use Simpson's Rule with n =

marta [7]

Answer:

The integral for the arc of length is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

By using Simpon’s rule we get: 1.5355453

And using technology we get: 2.3020

The approximation is about 33% smaller than the exact result.

Explanation:

The formula for the length of arc of the function f(x) in the interval [a,b] is:

$\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx$

We need the derivative of the function:

$f'(x)=\frac{1}{x}$

And we need it squared:

$[f'(x)]^2=\frac{1}{x^2}$

Then the integral is:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx$

Now, the Simposn’s rule with n=4 is:

$\displaystyle\int_a^b g(x)}dx\approx\frac{\Delta x}{3}\left( g(a)+4g(a+\Delta x)+2g(a+2\Delta x) +4g(a+3\Delta x)+g(b) \right)$

In this problem:

$a=1,b=3,n=4, \displaystyle\Delta x=\frac{b-a}{n}=\frac{2}{4}=\frac{1}{2},g(x)= \sqrt{1+\frac{1}{x^2}}$

So, the Simposn’s rule formula becomes:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\\\approx \frac{\frac{1}{3}}{3}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{1}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(1+\frac{2}{2}\right)^2}} +4\sqrt{1+\frac{1}{\left(1+\frac{3}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$

Then simplifying a bit:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx \approx \frac{1}{9}\left( \sqrt{1+\frac{1}{1^2}} +4\sqrt{1+\frac{1}{\left(\frac{3}{2}\right)^2}} +2\sqrt{1+\frac{1}{\left(2\right)^2}} +4\sqrt{1+\frac{1}{\left(\frac{5}{2}\right)^2}} +\sqrt{1+\frac{1}{3^2}} \right)$