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umka2103 [35]
3 years ago
12

The standard deviation of a sample taken from population A is 17.6 for a sample of 25. The standard deviation of a sample taken

from population B is 21.2 for a sample of 30.
Mathematics
1 answer:
agasfer [191]3 years ago
3 0
33.3..........................................
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What is 25.0996 rounded to the nearest thousandth
madreJ [45]

Answer:

25.100

Step-by-step explanation:

The thousandth place is the third digit to the right of the decimal point - tenths is the first, hundredths is the second, thousandths is the third.  Since six rounds up and both nines round up, one must round up all of the way from the fourth decimal place to the tenth decimal place.

4 0
3 years ago
If anyone knows about definite integrals for calculus then please I request help! I
kicyunya [14]

Answer:

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:                       \displaystyle du = \frac{-8}{x^3} \ dx
  3. [Bounds] Switch:                                                                                           \displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx
  2. [Integral] U-Substitution:                                                                              \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du
  3. [Integral] Exponential Integration:                                                               \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)
  5. Simplify:                                                                                                         \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

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

Unit: Integration

4 0
2 years ago
2. The direct distance between city A and city B is 200 miles. The direct distance
Misha Larkins [42]

Answer:

2

Step-by-step explanation:

8 0
3 years ago
HELP?! PLEASE! Which point is a reflection of7(-6.5,1) across the x-axis and y-axis?
Mandarinka [93]

Answer:

Dont need to worry

First, start off with the x-axis. -6.5, 1 becomes 6.5, 1. This is because point T is 6.5 to the left of the x-axis line, so our new point would be 6.5 to the right of the x-axis line. Same thing for the y-axis, (6.5, 1) would become (6.5, -1).

4 0
2 years ago
If point W is -4,-2 and point V is -2,-2 what is the length of VW
Helen [10]

Answer:

2 units

Step-by-step explanation:

The length of VW is just the distance between the two coordinates. You could use the distance formula, or an easier and faster way would be to recognize that since both points have the same y-coordinate, the distance between them will just be the distance of the x-coordinates. The absolute value of -4 and -2 is 2, so the length of VW is 2 units.

Hope this helped! ;)

5 0
3 years ago
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