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3 years ago

For which value(s) of x will the rational expression below be undefined?

Mathematics

1 answer:

zvonat [6]3 years ago

6 0

Question:

For which value(s) of x will the rational expression below be undefined? Check all that apply.

(x-3)(x+6)/x+7

Answer:

For x = -7 the rational expression is undefined

Solution:

Given rational expression is:

$\dfrac{(x-3)(x+6)}{(x+7)}$

We have to find the value of x for which the rational expression becomes undefined

A rational expression is undefined when the denominator is equal to zero

Here, the denominator becomes zero when x = -7

Substitute x = -7 in given

$\dfrac{(-7-3)(-7+6)}{(-7+7)} = \frac{-10 \times -1}{0} = \frac{10}{0}$

Thus for x = -7 the rational expression is undefined

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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

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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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