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2 years ago

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What is the NOT A RESTRICTION on the quotient of quantity 5 x squared minus 10 x minus 15 divided by quantity x squared minus 9

divided by quantity 6 x plus 12 divided by quantity x plus 2?
x ≠ -2
x ≠ 9
x ≠ 3
x ≠ -3

1 answer:

11111nata11111 [884]2 years ago

3 0

X not equal to 3

i hope this helps

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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.

<u>Start by calculating the mean</u>

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

<em>Hence, the standard deviation is approximately 11.28</em>

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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Answer:

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To solve for r, plug in the change in x values and change in y values, since the hypotenuse is simply the diagonal of both the y-coordinate and x-coordinates shown via drawing legs on the vertical and horizontal axis. So, since (0,0) is the initial point, r = sqr[(-4-0)^2 + (-3-0)^2 = sqr(16 + 9) = 5.

Now, the angle theta is the angle in which the sine, cosine, and tangent ratios are found. Simply use opposite over hypotenuse for sine, adjacent over hypotenuse for cosine, and opposite over adjacent for tangent using theta as the angle in which these values are obtained.

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Express: 0.006 in exponeutialform

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