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julia-pushkina [17]
3 years ago
10

PLEASEEE HELP MEEE DO THISSS♥️♥️♥️♥️​

Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0
Question 5 is C ASA but I did not know question 6
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Since Beth was born, the population of her town has increased at a rate of 850 people per year. On Beth's 9th birthday, the tota
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Answer:

editor edit

37,650

Step-by-step explanation:

37650

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Can someone tell me how to find my weighed grade. I have a 97% for homework which is 15% of my grade and 84% which is 35% of gra
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When sample size increases Group of answer choices
Olenka [21]

Answer:

Option A)

Confidence interval decreases

Step-by-step explanation:

If we increase the sample size, then,

  • The standard error of the interval decrease.
  • If the standard error increase, the margin of error of the interval decrease.
  • If the margin of error decreases, the width of the confidence level decreases, hence, the confidence interval become narrower.

Thus, the correct answer is

Option A)

Confidence interval decreases

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3 years ago
A right triangle has one vertex on the graph of y = 9 - x^2 , x > 0, at ( x , y ), another at the origin, and the third on th
jenyasd209 [6]

Answer:

  a.  A(x) = (1/2)x(9 -x^2)

  b.  x > 0 . . . or . . . 0 < x < 3 (see below)

  c.  A(2) = 5

  d.  x = √3; A(√3) = 3√3

Step-by-step explanation:

a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...

  A(x) = (1/2)(x)(y)

  A(x) = (1/2)(x)(9-x^2)

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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).

On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).

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c. A(2) = (1/2)(2)(9 -2^2) = 5

The area is 5 when x=2.

__

d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).

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How many seconds are in the month of February?
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the answer is 2,419,200 seconds

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