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3 years ago

Solve the given equation

Mathematics

1 answer:

AlexFokin [52]3 years ago

4 0

Eliminate the fractions by multiplying by least common multiple.
4,9,12 all go into 36 evenly. Multiply entire equation by 36.

36/4 = 9
36/9 = 4
36/12 = 3

New equation looks like:
$9(w-1) +4(w+1) = -3(w+1) \\ \\ 9w -9 +4w +4 = -3w -3 \\ \\ 9w +4w+3w = -3 +9 -4 \\ \\ 16w = 2 \\ \\ w = \frac{2}{16} = \frac{1}{8}$

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A circle has a circumference of 18. It has an arc of length 6. What is the central angle of the arc, in degrees?

nadya68 [22]

Circumference: 2πr C=18 so r= 18/2π= 9π

Arc length = rθ 6=(9π)θ θ=6/9π

Convert to degrees by multiplying by 180/π

6/9π x 180/π = 120 degrees

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3 years ago

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2. In your own words, explain HOW to find the standard deviation of this data set. (20 points) 3. What does the standard deviati

Ket [755]

Answer:

$\sigma = 11.28$ --- Standard deviation

Step-by-step explanation:

Given

See attachment for graph

Solving (a): Explain how the standard deviation is calculated.

Start by calculating the mean

To do this, we divide the sum of the products of grade and number of students by the total number of students;

i.e.

$\bar x = \frac{\sum fx}{\sum f}$

So, we have:

$\bar x = \frac{50 * 1 + 57 * 2 + 60 * 4 + 65 * 3 +72 *3 + 75 * 12 + 77 * 10 + 81 * 6 + 83 * 6 + 88 * 9 + 90 * 12 + 92 * 12 +95 * 2 + 99 * 4 + 100 * 5}{1 + 2 + 4 + 3 +3 + 12 + 10 + 6 + 6 + 9 + 12 + 12 +2 + 4 + 5}$

$\bar x = \frac{7531}{91}$

$\bar x = 82.76$

Next, calculate the variance using the following formula:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f}$

i.e subtract the mean from each dataset; take the squares; add up the squares; then divide the sum by the number of dataset

So, we have:

$\sigma^2 = \frac{1 * (50 - 82.76)^2 +...................+ 5 * (100 - 82.76)^2}{91}$

$\sigma^2 = \frac{11580.6816}{91}$

$\sigma^2 = 127.26$

Lastly, take the square root of the variance to get the standard deviation

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{127.26}$

$\sigma = 11.28$ --- approximated

Hence, the standard deviation is approximately 11.28

Considering the calculated mean (i.e. 82.76), the standard deviation (i.e. 11.28) is small and this means that the grade of the students are close to the average grade.

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