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3 years ago

Find the exact area of the region bounded by two concentric circles with radii 10 inches and 6 inches.

2 answers:

6 0

Area of circle
A = pi r^2
A = pi(10^2 - 6^2)
bounded by both is 36 pi
bounded by anyone is 100 pi
bounded by one only is 64 pi

Answer is: 64 PI

Hope that helps!!!

Hatshy [7]3 years ago

5 0

Answer: The area of the region bonded by two concentric circles with radii 10 inches and 6 inches is $64\pi$ in²

Step-by-step explanation:

Here, two circles having radius 10 inches and 6 inches are concentric,

Also, the area of the region bounded by these circles = Area of large circle - Area of small circle. ( Circle having greater radius is called large circle while having smaller area is called small circle )

= $\pi (10)^2 - \pi(6)^2$

= $\pi( 10^2 - 6^2)$

= $\pi ( 100 - 36 )$

= $64\pi\text{ square inches }$

Hence, The area of the region bonded by two concentric circles with radii 10 inches and 6 inches is $64\pi$ in²

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3 years ago

Read 2 more answers

Wayne Gretsky scored a Poisson mean six number of points per game. sixty percent of these were goals and forty percent were assi

BartSMP [9]

Answer:

a) The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

b) 0.05 = 5% probability that he has four goals and two assists in one game

Step-by-step explanation:

In hockey, a point is counted for each goal or assist of the player.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval. The standard deviation is the square root of the mean.

(a) Find the mean and standard deviation for the total revenue he earns per game.

60% of six are goals, which means that 60% of the time he earned 3K.

40% of six are goals, which means that 40% of the time he earned 1K.

The mean is:

$\mu = 6*0.6*3 + 6*0.4*1 = 13.2$

The standard deviation is:

$\sigma = \sqrt{\mu} = \sqrt{13.2} = 3.63$

The mean for the total revenue he earns per game is of 13.2K while the standard deviation is of 3.63K.

(b) What is the probability that he has four goals and two assists in one game

Goals and assists are independent of each other, which means that we find the probability P(A) of scoring four goals, the probability P(B) of getting two assists, and multiply them.

Probability of four goals:

60% of 6 are goals, which means that:

$\mu = 6*0.6 = 3.6$