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zepelin [54]
1 year ago
7

Set up an algebraic equation and then solve.

Mathematics
1 answer:
svetlana [45]1 year ago
5 0

Answer:

Width: 6

Length: 20

Step-by-step explanation:

So the area of a rectangle can be defined as: A=wl where w=width and l=length.

In this case we don't know what the length is, so let's just say the length is the variable l, and since the width is 14 units less than the length, we can express it as (l-14). this gives us the equation: A=l(l-14)=l^2-14l. We can solve for l, since we're given the area which is 120. So let's set the equation equal to that:

Original Equation:

A=l^2-14l

Substitute 120 as A (given)

120=l^2-14l

There is many ways to solve this equation: factoring, quadratic equation, completing the square etc... but in this case I'll just complete the square

Add (b/2)^2 to both sides to complete the square

120+(\frac{-14}{2})^2=l^2-14l+(\frac{-14}{2})^2

Simplify

169=l^2-14l+49

Rewrite right side a square binomial

169=(l-7)^2

Take the square root of both sides

13=l-7\\

Add 7 to both sides

20=l

to solve for width simply subtract 14 from the length which is 20, so the width is 6

Width: 6

L: 20

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My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the
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Answer:

a) n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

And rounded up we have that n=1476

b) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The margin of error for the proportion interval is given by this formula:  

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

If solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.9=0.1 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.64  

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.02 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

And rounded up we have that n=1476

Part b

For this case since we don't have a prior estimate we can use \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

8 0
3 years ago
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