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garik1379 [7]
3 years ago
6

Which transformations are needed to change the parent cosine function to the cosine function below?

Mathematics
2 answers:
Otrada [13]3 years ago
5 0

Question:

1. vertical stretch of 2, horizontal stretch to a period of 2 pi, phase shift of StartFraction pi Over 2 EndFraction units to the right, vertical shift of 1 unit up

2. vertical compression of 2, horizontal stretch to a period of StartFraction pi Over 4 EndFraction, phase shift of StartFraction pi Over 4 EndFraction units to the left, vertical shift of 2 units down

3. vertical compression of 2, horizontal compression to a period of 4 pi, phase shift of StartFraction pi Over 3 EndFraction units to the left, vertical shift of 1 unit down

4. vertical stretch of 2, horizontal compression to a period of StartFraction pi Over 2 EndFraction, phase shift of Pi units to the right, vertical shift of 1 unit down

Answer:

The correct option is;

4. Vertical stretch of 2, horizontal compression to a period of π/2, phase shift of π units to the right, vertical shift of 1 unit down

Step-by-step explanation:

Here, we have the parent cosine function given by y = cos(x)

A plot of the above function is attached

By comparison, it is observed that the transformation required to change the parent function includes;

1. Vertical stretch from the parent function of amplitude 1 to amplitude 2

2. A horizontal compression so that the period of one complete revolution is π/2 rather than 2·π in the parent cosine function

3. Phase shift of pi units to the right such that cos(π) = 1 rather than cos(2π) = 1 as it applies to the parent function

4. Shift of the y values from the symmetrical x-axis origin downwards by 1 unit such that the maximum y value of the doubly stretched transformation is 1 rather than 2

notsponge [240]3 years ago
4 0

Answer:

D)

Step-by-step explanation:

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(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot
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Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

a)

We could model this situation with a binomial distribution

P(6;k)=\binom{6}{k}p^kq^{6-k}

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

So

\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}

b)  

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}

P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281

P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484

P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438

P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109

P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781

and adding up these values we have that the probability that Y is less than 6 is

\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%

c)

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.  

In a binomial distribution, the standard deviation is  

s = \sqrt{npq}

In this case  

\sqrt{200(0.85)(0.15)}=5.05

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is  

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

np\geq 5 \;and\;nq \geq 5

in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30

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Answer:

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Step-by-step explanation:

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Answer:

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