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WITCHER [35]
3 years ago
9

The lines whose equations are 2x + y = 3z and x + y = 6z intersect at which point?

Mathematics
1 answer:
Tema [17]3 years ago
3 0
2x + y - (x + y) = 3z - 6z 
--------------------------------------------------------
<span>x = -3z </span>
------------------------------------------

<span>(-3z, 9z) </span>
 ----------------------------------
<span>(x,y,z) = (-3z, 9z, z) 
</span>--------------------------
<span>x = -3z, y = 9z
---------------------
There for the answer is </span><span>(-3z, 9z) </span>
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888 divided by 8 and<br>574 divided by 7<br>​
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Answer:

111 & 82

Step-by-step explanation:

888/8=111

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Sam needs to cut a piece of sheet metal into 8 pieces. It takes him 5minutes to make each cut. How many cuts will sam make?
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In triangle ABC, a = 9, c = 5, and B = 120°. Find b2.
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Fewer young people are driving. In 1983, 87% of 19-year-olds had a driver’s license. Twenty-five years later (in 2008) that perc
Dima020 [189]

Answer:

a) ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019  

b) ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245  

c) On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of \hat p the margin of error change for part a and b.

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.  

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 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=\pm 1.96

If we replace the values into equation (a) for 1983 we got:

ME=1.96\sqrt{\frac{0.87 (1-0.87)}{1200}}=0.019  

Part b

Since is the same confidence level the z value it's the same.  

If we replace the values into equation (a) for 2008 we got:

ME=1.96\sqrt{\frac{0.75 (1-0.75)}{1200}}=0.0245  

Is the margin of error the same in parts (a) and (b)? Why or why not?

On this case it's not the same since the proportion estimated for 1983 it's different from the proportion estimated for 2008. So since the margin of error depends of \hat p the margin of error change for part a and b.

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lawyer [7]

Answer:

More students have hazel eyes than blue.

Step-by-step explanation:

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