All categories

3 years ago

Find the area of a circle with diameter, d = 18m.Give your answer in terms of π.

Mathematics

2 answers:

Fofino [41]3 years ago

8 0

Answer:

81π

Step-by-step explanation:

Area of a circle formula is : $A=\pi r^2$ . r stands for radius.

Radius is half of the diameter. To solve for the area of the circle, we must replace r with diameter/2 which is 9 (18/2 = 9).

$A=\pi 9^{2}$

Now, the question wants us to give the answer in terms of pi. So we are going to simplify the equation WITH PI. Meaning, we aren't going to multiply pi. So we will just simplify 9 squared to 81.

So in terms of pi the area is 81π

Svetllana [295]3 years ago

6 0

Answer:

81π

Step-by-step explanation:

Area of a circle= π $r^{2}$ ( where r is the radius multiplying it twice)

Area of a circle = π × r × r ( this is the formula in simplier terms )

Area of a circle = 3.14 × 9 × 9 ( To get 9 divide 18 by 2 as diameter is 2 radii

Area of a circle = 81π ( multiply the numbers AND THESE ARE THE TWO NINES LEAVING π)

HOPE IT HELPED

You might be interested in

Value of x is 3.2x-9.6=38.4?

Svetradugi [14.3K]

Answer:

The steps that can be done to both sides of the equation is: Add 9.6, then divide by 3.2

Step-by-step explanation:

Solve the equation:

$3.2x - 9.6 = 38.4$

-First you add both sides 9.6:

$3.2x -9.6 +9.6 = 38.4 +9.6$

$3.2x = 48$

-Then, you divide both sides by 3.2:

$\frac{3.2x}{3.2} = \frac{48}{3.2}$

$x=15$

7 0

3 years ago

125+[-125]=? PLEASE HELP?!

elena55 [62]

The answer to this question is 0.

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Using number line what is 4-5

Vinil7 [7]

Answer:

Step-by-step explanation:

4 - 5 = -1

Count:

4, 3, 2, 1, 0, -1

6 0

3 years ago

How do you write (6)^-2 without exponents?

m_a_m_a [10]

(6)^-2 Is equivalent to 1/36

5 0

3 years ago

Read 2 more answers

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

Find the area of a circle with diameter, d = 18m.Give your answer in terms of π.​

Find the area of a circle with diameter, d = 18m.Give your answer in terms of π.