Simplify, picture attached

The face of a clock has a circumference of 63 in. What is the area of the face of the clock?

ryzh [129]

Answer:

The area of the clock $= 315.41\ inch^{2}$

Step-by-step explanation:

We have been given the face of the clock that is $63\ in$

So that is also the circumference of the clock.

Since the clock is circular in shape.

So $2\pi(r)=63\ inch$

From here we will calculate the value of radius $(r)$ of the clock that is circular in shape.

Then $2\pi(r)=63\ inch =\frac{63}{2\pi} = 10.02\ in$

Now to find the area of the clock we will put this value of (r) in the equation of area of the circle.

Now $\pi (r)^{2}=\pi(10.02)^{2}=315.41\ in^{2}$

So the area of the face of the clock = $315.41\ in^{2}$

6 0

2 years ago

Two angles are supplementary. The measure

BlackZzzverrR [31]

Answer:

57°

Step-by-step explanation:

Let's assume the measurement of the second angle is x.

so, following the problem

(2x + 9) + x = 180

2x + x = 3x

=> 3x + 9 = 180

=> 3x = 180 - 9

3x = 171

x = 171/3

x = 57°

57° is the smallest angle

5 0

2 years ago

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> $\mu$ = [ $\bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } }$ ]

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

= [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0

2 years ago

Jen measured the growth of a sunflower in one week it grew 2 and 1 half in

Pie

So you basically add the amount it grew each week. So week one is 2 and 1/2, week two is 2 and 3/4 and week three is 3 and 1/4. to make things simpler, add week two and three which gets you 6inches, then add week one, to get 8 and a half inches.

8 0

3 years ago

Find the equation of the line that passes through the following two points: (3, -7) and (7, 2)

Wittaler [7]

Answer:

$y=\frac{9}{4}x-\frac{55}{4}$