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3 years ago

What is the area of a circle with inner radius of 2 cm and outer radius is 4 cm

Mathematics

1 answer:

anygoal [31]3 years ago

6 0

the area of a circle is given for the equation A = pi*r^2,
for a radius of 2 cm , A = 3,1416*(2)^2 = 12,56 CM2
For a radius of 4 cm A = 3,1416* (4)2= 50,26 CM2

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What is the value of -17 - (-1) + 2/3 + -1/2?

netineya [11]

Answer:

-95/6 or -15.83 repeating

Step-by-step explanation:

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3 years ago

First answer I will give you brainleyist

asambeis [7]

Answer: C 6 cm
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2 years ago

A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its op

Alina [70]

Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, $\dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h$

Volume of cone is given by

$V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3$

By differentiating with respect to time t, we get that

$\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}$

Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie $\dfrac{dv}{dt}=-4\ in^3$

and h = 2 inches deep.

$-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}$

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

7 0

2 years ago

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the

Vesna [10]

Answer:

A sample size of at least 1,353,733 is required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

You would like to be 98% confident that you esimate is within 0.1% of the true population proportion. How large of a sample size is required?

We need a sample size of at least n.

n is found when M = 0.001.

Since we don't have an estimate for the proportion, we use the worst case scenario, that is $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.001 = 2.327\sqrt{\frac{0.5*0.5}{n}}$

$0.001\sqrt{n} = 2.327*0.5$

$\sqrt{n} = \frac{2.327*0.5}{0.001}$

$(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.001})^{2}$

$n = 1353732.25$

Rounding up

A sample size of at least 1,353,733 is required.

5 0

3 years ago

Find the equation of the line that passes through (-1, 4) and is perpendicular to the

statuscvo [17]

For perpendicular lines, m2 = -1/m1; where m1 is the slope of line 1 and m2 is the slope of line 2.
m1 = (-4 - 2)/(4 - 2) = -6/2 = -3
m2 = -1/-3 = 1/3
Equation of the required line is given by: y - 4 = 1/3 (x - (-1))
y - 4 = 1/3 x + 1/3
y = 1/3 x + 1/3 + 4
y = 1/3 x + 13/3

6 0

2 years ago