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oksano4ka [1.4K]
3 years ago
9

What value of x is in the solution set of 2x – 3 > 11 – 5x?

Mathematics
2 answers:
ohaa [14]3 years ago
4 0
Add three to both sides
2x greater than 14 - 5x
Add 5x to both sides
7x greater than 14
divide both sides by 7
x is greater than 2
Umnica [9.8K]3 years ago
3 0

Answer:

Thus, for x > 2  is in the solution set of 2x – 3 > 11 – 5x

Step-by-step explanation:

Given inequality  2x – 3 > 11 – 5x

We have to find the value of x is in the solution set.

We solve the given inequality,

Consider inequality 2x – 3 > 11 – 5x

Add 3 both side, we get,

2x – 3 + 3 > 11 – 5x + 3

On simplifying, we get,

2x > 14 - 5x

Now, add 5x both side, we get

2x + 5x > 14 - 5x + 5x

7x > 14

Divide both side by7, we get

x > 2

Thus, for x > 2  is in the solution set of 2x – 3 > 11 – 5x

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lower limit =0.261

upper limit =0.369

Step-by-step explanation:

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.  

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Description in words of the parameter p

p represent the real population proportion of people that have experienced bullying

X= 63 people in the random sample that have experienced bullying

n=200 is the sample size required  

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that have experienced bullying

z_{\alpha/2} represent the critical value for the margin of error  

The population proportion have the following distribution  

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Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{63}{200}=0.315 represent the estimated proportion of people that they were planning to pursue a graduate degree

Confidence interval

The confidence interval for a proportion is given by this formula  

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

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:  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.261  

0.315 - 1.64 \sqrt{\frac{0.315(1-0.315)}{200}}=0.369  

And the 90% confidence interval would be given (0.261;0.369).  

We are confident at 90% that the true proportion of people that they were planning to pursue a graduate degree is between (0.261;0.369).  

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