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AnnZ [28]
3 years ago
11

The circumference of the circle is increasing at a rate of 0.5 meters per minute. What's the rate of change of the area of the c

ircle when the radius is 4 meters?
1: 3 meters per minute
2: 4 meters squared per minute
3: 4 meters per minute
4: 2 meters squared per minute
5: 7 meters per minute

This is actually for a game but I'm really bad at math.
Mathematics
1 answer:
morpeh [17]3 years ago
3 0

Answer:

The rate of change of the area of the circle when the radius is 4 meters = 2 meters²/minute ⇒ answer 4

Step-by-step explanation:

* Lets revise the chain rule in the derivative

- If dy/da = m and dx/da = n, and you want to find dy/dx

∴ dy/dx = dy/da ÷ dx/da = m ÷ n = m/n

* In our problem we have

- The rate of increasing of the circumference dC/dt = 0.5 meters/minute

- We need the find the rate of change of the area of the circle

 when the radius is 4 meters

- The common element between the circumference and the area

 of the circle is the radius of the circle

* We must to find dC/dr and dA/dr and use the chain rule to

 find dA/dr

- Find the rate of change of the radius dr/dt

∵ C = 2πr

- Find the derivative of C with respect to r

∴ dC/dr = 2π ⇒ (1)

∵ dC/dt = 0.5 meters/minute ⇒ (2)

- Divide (1) by (2) to get dr/dt by using chain rule

∵ dC/dt ÷ dC/dr = 0.5 ÷ 2π

∴ dC/dt × dr/dC = 0.5 × 1/2π ⇒ cancel dC together and change

   0.5 to 1/2

∴ dr/dt = 1/2 × 1/2π = 1/4π ⇒ (3)

- Find the rate of change of the area dA/dt

∵ A = πr²

- Find the derivative of A with respect to r

∴ dA/dr = 2πr

∵ r = 4

∴ dA/dr = 2π(4) = 8π ⇒ (4)

- Multiply (4) by (3) to get dA/dt by using chain rule

∵ dA/dr × dr/dt = 8π × 1/4π ⇒ divide 8 by 4 and cancel π

∴ dA/dt = 2 meters²/minute

* The rate of change of the area of the circle when the radius is

  4 meters = 2 meters²/minute

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Katyanochek1 [597]

Answer:

a. 0.2898

b. 0.0218

c. 0.1210

d. 0.1515

e. This is because the population is normally distributed.

Step-by-step explanation:

Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal places

We are using the z score formula when random samples

This is given as:

z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation.

n is the random number of samples

a.If 100 SAT scores are randomly selected, find the probability that they have a mean less than 1500.

For x = 1500, n = 100

z = 1500 - 1518/325/√100

z = -18/325/10

z = -18/32.5

z = -0.55385

Probability value from Z-Table:

P(x<1500) = 0.28984

Approximately = 0.2898

b. If 64 SAT scores are randomly selected, find the probability that they have a mean greater than 1600

For x = 1600, n = 64

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z= 1600 - 1518 /325/8

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Probability value from Z-Table:

P(x<1600) = 0.97823

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c. If 25 SAT scores are randomly selected, find the probability that they have a mean between 1550 and 1575

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z = 1550 - 1518/325/√25

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z = 1550 - 1518/65

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Probability value from Z-Table:

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For x = 1575 , n = 25

z = 1575 - 1518/325/√25

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z = 1575 - 1518/65

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Probability value from Z-Table:

P(x=1575) = 0.80974

The probability that they have a mean between 1550 and 1575

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d. If 16 SAT scores are randomly selected, find the probability that they have a mean between 1440 and 1480

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z = 1440 - 1518/325/√16

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For x = 1480, n = 16

z = 1480 - 1518/325/√16

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Probability value from Z-Table:

P(x = 1480) = 0.32

The probability that they have a mean between 1440 and 1480

P(x = 1480) - P(x = 1440)

= 0.32 - 0.16853

= 0.15147

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e. In part c and part d, why can the central limit theorem be used even though the sample size does not exceed 30?

The central theorem can be used even though the sample size does not exceed 30 because the population is normally distributed.

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