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3 years ago

If x+y=12,and xy =22,find x^2+y^2

Mathematics

2 answers:

Pepsi [2]3 years ago

6 0

Here,

x+y=12 and xy=22

$\tt{(x+y)^2=x^2+y^2+2xy }$ ⠀

according to the question,

$\tt{ (12)^2=x^2+y^2+2×22 }$ ⠀

$\tt{ x^2+y^2=144-44 }$ ⠀

$\tt{x^2+y^2=100 }$ ⠀

DochEvi [55]3 years ago

6 0

Answer:

$\bold{ x^2+y^2=100 }$

Step-by-step explanation:

In questions,

x+y=12

xy=22

so,

$\mapsto{x^2+y^2+2ab=(x+y)^2 }$

$\mapsto{x^2+y^2+2×22=144 }$

$\mapsto{x^2+y^2=144-44 }$

$\mapsto{ x^2+y^2=100 }$

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A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

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$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

3 0

3 years ago

If x+y=12,and xy =22,find x^2+y^2​

If x+y=12,and xy =22,find x^2+y^2