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AnnyKZ [126]
3 years ago
5

Write an equation for the parent function y = x^ 2 stretched by a factor of 6 and shifted down 4 units?

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0

Answer:

 

Translations

y = f (x) + k up k units

y = f (x) - k down k units

y = f (x + h) left h units

y = f (x - h) right h units

Stretches/Shrinks

y = m·f (x) stretch vertically by a factor of m

y = ·f (x) shrink vertically by a factor of m (stretch by  

y = f (x) stretch horizonally by a factor of n

y = f (nx) shrink horizontally by a factor of n (stretch by )

Reflections

y = - f (x) reflect over x-axis (over line y = 0)

y = f (- x) reflect over y-axis (over line x = 0)

x = f (y) reflect over line y = x

Hope this helps

Step-by-step explanation:

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a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

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Step-by-step explanation:

Data given and notation  

n=2700 represent the random sample taken    

X represent the people in England who are deficient in Vitamin D

\hat p=0.2 estimated proportion of England people who are deficient in Vitamin D

\alpha=0.02 represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)    

p= population proportion of England people who are deficient in Vitamin D

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

And replacing into the confidence interval formula we got:

0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182

0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

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