Divide -0.0328 0.002 What is the quotient? Enter your answer in the box.

Mathematics

1 answer:

Marina86 [1]1 year ago

7 0

The quotient is -16.4

$\frac{-0.0328}{0.002}$

Apply the fraction rule: $\frac{-a}{b}=-\frac{a}{b}$ $=-\frac{0.0328}{0.002}$

Divide the numbers: $\frac{0.0328}{0.002}=16.4$ $=-16.4$

The quotient rule is a calculus technique for determining the derivative or differentiation of a function that is represented as a division or ratio of two differentiable functions.

Therefore, we can apply the quotient method to get the derivative of a function of the form f(x)/g(x), where f(x) and g(x) are both differentiable functions and g(x) ≠ 0. The quotient rule immediately follows the product rule and the notion of limits of derivation in differentiation.

Learn more about quotient rule brainly.com/question/15568137

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There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s

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Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

Original Mean = $70,000

$\frac{\sum X}{n}$ = $70,000 ⇒ $\sum X$ = 70,000 * 10 = $700,000

New $\sum X$ after change = $700,000 - $100,000 + $1,000,000 = $1600000

Therefore, New mean = $\frac{1600000}{10}$ = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = $20000^{2}$ i.e. $20000^{2}$ = $\frac{\sum x^{2} - n*xbar }{n -1}$

Original $\sum x^{2}$ = ( $20000^{2}$ * (10-1)) + (10 * 70,000) = $3,600,700,000

New $\sum x^{2}$ = $3,600,700,000 - $100,000^{2} + 1,000,000^{2}$ = 9.936007 * $10^{11}$

New Variance = $\frac{new\sum x^{2} - n*new xbar }{n -1}$ = $\frac{9.936007 *10^{11} - 10*160000 }{10 -1}$ = 1.103999 * $10^{11}$ Therefore, standard deviation after change = $\sqrt{1.103999 * 10^{11} }$ = $332264.804 .