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Oksanka [162]
3 years ago
11

What's the answer to 4|p-3|=|2p+8|? And how do you check your answer?

Mathematics
1 answer:
Serhud [2]3 years ago
4 0
We want to solve 4|p-3| = |2p+8|

Ths equation is equvalent to 4 equatons.

Case 1.
4(p - 3) = 2p + 8
4p - 12 = 2p + 8
2p = 20
Answer:  p = 10

Check the solution. 
4|p-3| = 4*7 = 28
|2p+8| = 2*10+8 = 28
The solution is correct.

Case 2.
4(3 - p) = 2p + 8
12 - 4p = 2p + 8
4 = 6p
Answer:  p = 2/3

Check the solution.
4|p-3| = 4*|(2/3-3)| = 28/3
|2p+8| = 28/3
The solution is correct.

Case 3.
4(p-3) = -2p - 8
4p - 12 = -2p - 8
6p = 4
Answer: p = 2/3
The solution is correct because it was checked in case 2.

Case 4.
4(3 - p) = -2p - 8
12 - 4p = -2p - 8
20 = 2p
Answer:  p = 10
The answer s correct because t was checked in Case 1.

Answers:  
p = 10 or  p = 2/3.  
Both answers check out to be correct.

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3 years ago
My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the
AVprozaik [17]

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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