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4 years ago

Can someone help me with this geometry problem please ?

Mathematics

1 answer:

Oxana [17]4 years ago

5 0

Answer:

P and Q

Step-by-step explanation:

In geometry, a transversal is a line that passes through two lines in the same plane at two distinct points. In most cases there is only one but here there are two present. Transversals play a role in establishing whether two other lines in the Euclidean plane are parallel.

Hope this helps!

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If f(x)=x^1/2 and g(x)=2x^3-x^1/2, find f(x)-g(x)... HELP ITS URGENT

bija089 [108]

F(x) - g(x) =
x^1/2 - (2x^3 - x^1/2)=
x^1/2 - 2x^3 + x^1/2=
2x^1/2 - 2x^3

7 0

4 years ago

Diego has 3 times as many comic books as Han. How many comic books does Diego have if Han has: 6 comic books? N books? Diego has

Annette [7]

Answer:

the answer is 9

Step-by-step explanation:

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3 years ago

A punch makes 16 pints.If 10 guests equally share the punch,how many pints will each guest drink?

m_a_m_a [10]

Answer:

1.6 pints

Step-by-step explanation:

A punch=16 pints

If 10 guest shares the punch equally

Each guest get=16 pints/10

=1.6 pints

Each guest will drink 1.6 pints of punch

8 0

3 years ago

Help me please.<br> I have 2 attachments so you can see the graphs.

KIM [24]

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5 0

3 years ago

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

pav-90 [236]

Answer:

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

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