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harkovskaia [24]
3 years ago
10

The graph of a quadratic function has a vertex located at (7,-3) and passes through points (5,5) and (9,5). Which equation best

represents this function?
A) f(x)=(x-7)^2-3
B) f(x)=2(x-7)^2-3
C) f(x)= -(x-5)^2+5
D) f(x)= -2(x-5)^2+5
Mathematics
1 answer:
Serga [27]3 years ago
4 0

Answer:

B) f(x)=2(x-7)^2-3.

Step-by-step explanation:

The vertex form of the equation is given by f(x)=a(x-h)^2+k.

We plug in the vertex to obtain:

f(x)=a(x-7)^2-3.

Since the graph passes through (5,5) and (9,5), they must satisfy its equation.

5=a(9-7)^2-3.

5+3=4a.

8=4a

Divide both sides by 4.

a=2

Therefore the equation is:

f(x)=2(x-7)^2-3.

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Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak
crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28

We can now calculate the probabilities:

P(0.47

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2

We can now calculate the probabilities:

P(0.47

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

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