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kicyunya [14]
3 years ago
12

Solve for x and y. they both have to be the same number for each equation

Mathematics
1 answer:
Elis [28]3 years ago
4 0
Hey there! :D

y=2x-9

We can plug that in as y, since y is equal to 2x-9. 

5x-3(2x-9)= 23

Distribute. 

5x-6x+27= 23

-1x+27=23

Subtract 27 on both sides. 

-1x=-4

Divide both sides by -1.

x=4

Now, plug that in for y. 

y=2(4)-9

y=8-9

y=-1

 (4,-1) <== solution

I hope this helps!
~kaikers
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Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr
Minchanka [31]

Answer:

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98[/tex[ or 98%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".  [tex]p_1 represent the real population proportion for 1

\hat p_1 =0.768 represent the estimated proportion for 1

n_1=92 is the sample size required for 1

p_2 represent the real population proportion for 2

\hat p_2 =0.646 represent the estimated proportion for 2

n_2=95 is the sample size required for 2

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_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}  

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

ME= \frac{0.2753-(-0.0313)}{2}=0.1533

And we know that the margin of erro is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}

We have all the values except the value for z_{\alpha/2}

So we can find it like this:

0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}

And solving for z_{\alpha/2} we got:

z_{\alpha/2}=2.326

And we can find the value for \alpha/2 with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got \alpha/2 =0.01 so then the value for \alpha=0.02 and then the confidence level is given by: Conf=1-0.02=0.98 or 98%

7 0
3 years ago
Can someone help me with 42 divided by 9
Gnom [1K]

Answer:

4.6 or if you rounded it would be 5

8 0
3 years ago
The average age of instructors at a local college is 55, with a standard deviation of 6 years; the average salary is $37,000 wit
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48/80 divided by what equals 3/5 ?
mote1985 [20]
They have the exact same value so you just divide by 1
6 0
3 years ago
Read 2 more answers
Answers please please please with a cherry on top!
ipn [44]
Would you like a full explanation or is it better for me just to plug it all in to my graphing calculator to get quick answers
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