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2 years ago

What is the mean of this data? 7,5,5,3,2,2

Mathematics

1 answer:

Vikentia [17]2 years ago

8 0

Answer:

Step-by-step explanation:

The mean is the average of a data set. It can be found by adding up all of the values in a data set and then dividing it by the number of values in the data set.

The values in this data set;

$7,5,5,3,2,2$

The number of values in this data set,

$6$

Find the mean;

$\frac{sum\ of\ vlaues}{number\ of\ values}$

$=\frac{7+5+5+3+2+2}{6}\\\\=\frac{24}{6}\\\\=4$

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Answer:

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2 years ago

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The classrooms are trying to raise $1,200.00 for charity.

Svetradugi [14.3K]

Answer:

The classes need to raise $164.86 more to reach their goal of $1200.

Step-by-step explanation:

$1,200 total

Mrs. LeBlanc's class has raised 50% of the total, so they've raised $600.

1200/2 = 600

Mr. Patel's class has raised $235.14.

235.14 + 600 = 835.14

Ms. Warner's class has raised 1/3 of the total of Mrs. LeBlanc's class ($600), so they've raised $200.

600/3 = 200

835.14 + 200 = 1035.14

The three classes have raised a combined total of $1035.14.

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3 years ago

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antiseptic1488 [7]

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2 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:

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

Then we just do those computations and we finally get the approximation via Simposn's rule:

$\displaystyle\int_1^3\sqrt{1+\frac{1}{x^2}}dx\approx 1.5355453$

While when we do the integral by using technology we get: 2.3020.

The approximation with Simpon’s rule is close but about 33% smaller:

$\displaystyle\frac{2.3020-1.5355453}{2.3020}\cdot100\%\approx 33\%$

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2 years ago

The temperature yesterday was 11 degrees colder than it was today if she represents the temperature today what expression repres

damaskus [11]

68-11=57 (Yesterday's temperature)
That's the expression. ^^ And yesterday's temperature is 57 because, 68 degrees (Today's temperature) minus 11 degrees (how many degrees colder it was yesterday) is equal to 57 degrees (Yesterday's Temperature).

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3 years ago