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2 years ago

A circle has a diameter with endpoints at (0, 4) and (0, –8). What is the approximate area of the circle?

Mathematics

2 answers:

Morgarella [4.7K]2 years ago

6 0

Answer:

$36\pi$

Step-by-step explanation:

The diameter is:

The Euclidean distance between points (x1, y1) and (x2, y2) is:

sqrt((x1 - x2)^2 + (y1 - y2)^2)

Substitute (x1, y1) = (0, 4) and (x2, y2) = (0, -8):

= sqrt((-0 + 0)^2 + (-(-8) + 4)^2)

= sqrt(12^2)

= 12

or simply:

diameter: 4-(-8)=12

The radius is:

12/2=6 units

The area of the circle:

$\pi r^2=\pi(6^2)=36\pi$

nadya68 [22]2 years ago

4 0

Answer:

113.04

Step-by-step explanation:

Area of a circle is πr2

π = 3.14

r = 6

πr2 = 113.04

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The diagonals of a rhombus are 6 inches and 8 inches. What is the perimeter of the rhombus?

loris [4]

The perimeter of the rhombus is 28 inches.

4 0

2 years ago

Read 2 more answers

A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we

weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weighing of specimen = $\frac{3.412+3.416+3.414}{3}$ = 3.414 g

$\sigma$ = population standard deviation = 0.001 g

n = sample of specimen = 3

$\mu$ = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

<u>99% confidence interval for</u> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } }$ , $3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } }$ ]