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2 years ago

The outlier, 78, is removed removed from the data set shown

Mathematics

2 answers:

Dvinal [7]2 years ago

7 0

Minus the orignal range 55 from the new range 22

d. 55-22
=33

Elanso [62]2 years ago

5 0

The range is decreased by 33.

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rodikova [14]

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2 years ago

Need help!!! which addition matches the problem...

Tom [10]

Answer:

The addition that matches the problem is the first one.

Step-by-step explanation:

The red line goes back 8 so it is a -8 and the blue line goes forward 4 so it is a postive 4.

4 + (-8) = -4

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2 years ago

Evaluate -3x+4y when x=-7 y=-10

Alexus [3.1K]

Answer:

-19

Step-by-step explanation:

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

7 divided by 20 is what answers

Montano1993 [528]

Hello!

The Correct Answer to this is that 7 divided by 20 or 7/20 equals:

"7/20 = 0.35"

Explanation:

Since you are trying to find equivalent values for 7/20, you can make two proportions and set them equal to each other. The following states that "7 out of 20 is equal to some amount out of 100."

7/20=x/100

Solve by cross multiplying:

20x=700

Divide both sides by 20 to isolate x:

x=35 Therefore, 7/20=35/100. This is the same as saying 35%, since by definition "per" means out of, and "cent" means hundred. To make it into a decimal just move the decimal place two digits to the left, such that 35.00 becomes 0.35, and 100.00 becomes 1. Then it is simply 0.35/1, or 0.35

Hope this Helps! Have A Wonderful Day! :)

5 0

2 years ago

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