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3 years ago

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Please help :,). For the figures below, assume they are made of semicircles, quarter circles, and squares. For each shape, find

the area and perimeter. Give your answer as a completely simplified exact value in terms of π
(no approximations).

1 answer:

ad-work [718]3 years ago

8 0

Answer:

Area is 24^2 = 576.

Perimeter is 144 pi + 48

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The point P(x, y) lies on the terminal side of an angle mc014-1.jpg = –60° in standard position. What are the signs of the value

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<span>The sign value for x is positive and for y is negative. In standard position, the x point is at the origin, so the value will be 0. The terminal side is the line that moves to create the angle. At -60 degrees, the terminal side will be in the minus side of the y axis.</span>

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Which situation is most likely to have a constant rate of change?

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The <u><em>correct answer</em></u> is:

B) The cost for a group of friends to see a movie compared with the number of friends

Explanation:

Movie tickets typically cost a specific amount per person. This means that every person you take would add that cost to the total; this would be a constant rate of change, as there would be the same difference between successive points.

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You have received an order of 100 robotic resistance spot welders which contains 5 defective welders. You randomly select 15 wel

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Answer:

a)

$P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357$

$P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034$

$P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377$

$P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216$

$P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015$

$P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004$

b) 0.154% probability that there are at least 4 defective welders in the sample

Step-by-step explanation:

The welders are chosen without replacement, so the hypergeometric distribution is used.

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

100 welders, so $N = 100$

Sample of 15, so $n = 15$

In total, 5 defective, so $k = 5$

(a) Determine the PMF of the number of defective welders in your sample?

There are 5 defective, so this is P(X = 0) to P(X = 5). Then

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357$

$P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034$

$P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377$

$P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216$

$P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015$

$P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004$

(b) Determine the probability that there are at least 4 defective welders in the sample?

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0015 + 0.00004 = 0.00154$

0.154% probability that there are at least 4 defective welders in the sample

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A boat's crew rowed 15 kilometers downstream, with the current, in 1 hours. The return trip upstream, against the current, cover

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B and c are the speeds of the boat (in still water) and the current.

Traveling with the current, the boat goes b+c km per hour.
1 hour * (b+c km)/hour = 15 km
b+c = 15

Traveling against the current, the boat goes b-c km per hour.
3 hours * (b-c km)/hour = 15
b-c = 5

Add the two equations
b = 10 km per hour

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3 years ago