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umka21 [38]
3 years ago
9

Solve |z| > (1/2)

Mathematics
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

{z|z}U{z|z>\frac{1}{2}}

Step-by-step explanation:

Given the inequality |z| > \frac{1}{2} you need to set up two posibilities:

FIRST POSIBILITY : z>\frac{1}{2}

SECOND POSIBILTY: z

Therefore, you got that:

z\frac{1}{2}

Knowing this, you can write the solution obtained in Set notation. This is:

Solution: {z|z}U{z|z>\frac{1}{2}}

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The measures of angle 1 and angle 2 are 30% and 45% of the sum of the angle measures of the triangle. Find the value of x. ASAP
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Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex
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Answer:

(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942  

(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between 0.0942 \leq p_A -p_B \leq 0.09978

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".  

p_A represent the real population proportion female for Biology

\hat p_A =\frac{84199}{144796}=0.582 represent the estimated proportion female for biology

n_A=144796 is the sample size for A

p_B represent the real population proportion female for calculus AB

\hat p_B =\frac{102598}{211693}=0.485 represent the estimated proportion female for Calculus AB

n_B=211693 is the sample size required for B

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})  

The confidence interval for the difference of two proportions would be given by this formula  

(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}  

For the 90% confidence interval the value of \alpha=1-0.90=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.64  

And replacing into the confidence interval formula we got:  

(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942  

(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978  

And the 90% confidence interval would be given (0.0942;0.09978).  

We are confident at 90% that the difference between the two proportions is between 0.0942 \leq p_A -p_B \leq 0.09978

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