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Rzqust [24]
4 years ago
6

How many teenagers should be in the sample

Mathematics
1 answer:
valentinak56 [21]4 years ago
4 0

Answer:

THEREFORE

Step-by-step explanation:

THERE SHOULD BE 7 TEENAGERS IN THE SAMPLE.

HOPE, THIS MAY HELP YOU BRO!!..

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Need rn! worth 14 ~ 謝謝
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First is 14.4

Second I’m not sure

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Given only the information in the diagram, determine if m is parallel to n . If so, name the postulate or theorem used to suppor
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From my research on the internet, the image attached supports this problem. The two lines are parallel, as supported by the converse of corresponding angles postulate. It states that:<span> If a </span>transversal<span> intersects two lines and the corresponding angles are </span>congruent<span>, then the lines are parallel.</span>

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Read 2 more answers
What is the answer to 5 - a0 = 14
nikklg [1K]

Answer:

There are no solutions.

Step-by-step explanation:

cuz they ain't all real numbers

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3 years ago
Arrange the numbers as they appear from left to right on a horizontal number line. -5/7, 2 5/7, 5/7, -1 5/7, 1 2/7, -1/7, -2 2/7
zhuklara [117]

Answer:

-2 2/7, -1 5/7, -5/7, -1/7, 5/7, 1 2/7, 2 5/7

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3 years ago
Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is re
Tomtit [17]

Answer:

<u>Volume = 1.535</u>

<u />

Step-by-step explanation:

The region R is bounded by the equations:

y = √sin⁻¹x

y = √(π/2)

y = √(π/3)

x = 0

R is revolved around the x-axis so we will need f(y) for finding out the volume. We need to make x the subject of the equation and then replace it with f(y).

f(x) = √sin⁻¹x

y = √sin⁻¹x

Squaring both sides we get:

y² = sin⁻¹x

x = sin (y²)

f(y) = sin (y²)

Using the Shell Method to find the volume of the solid when R is revolved around the x-axis:

V = 2\pi \int\limits^a_b {f(y)} \, dy

The limits a and b are the equations y = √(π/2) and y = √(π/3) which bound the region R. So, a = √(π/2) and b = √(π/3).

V = 2π \int\limits^\sqrt{\frac{\pi }{2}}_\sqrt{\frac{\pi }{3} }   sin (y²) dy

Integrating sin (y²) dy, we get:

-cos(y²)/2y

So,

V = 2π [-cos(y²)/2y] with limits √(π/2) and √(π/3)

V = 2π [(-cos(√(π/2) ²)/2*√(π/2)] - [(-cos(√(π/3) ²)/2*√(π/3)]

V = 2π [(-cos(π/2)/ 2√(π/2)) - ((-cos(π/3)/ 2√(π/3))]

V = 2π [ 0 - (-0.5/2.0466)]

V = 2π (0.2443)

V = 1.53499 ≅ 1.535

3 0
3 years ago
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