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3 years ago

3(x -2) = 12 how to solve and what's the answer ?

Mathematics

2 answers:

Kamila [148]3 years ago

6 0

3(x - 2) = 12

Distribute 3 inside the parentheses.

3x - 6 = 12

Add 6 to both sides.

3x = 18

Divide both sides by 3.

$\boxed {x =6}$

AlexFokin [52]3 years ago

5 0

3(x-2)=12

Distribute :
3x-6=12

Add 6 to both sides to isolate the x variable
3x-6=12
+6. +6
3x=18

Divide by 3
(3x=18)÷3

X=6

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b) 90% of all samples of size 10 have sample means within 0.5035 of the population mean.

c) The 90% confidence interval would be given by (6.636;7.643)

d) Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

e) If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

Step-by-step explanation:

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".

$\bar X=7.14$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.87 represent the sample standard deviation

n=10 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that $t_{\alpha/2}=1.83$

$ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

Part b

90% of all samples of size 10 have sample means within 0.5035 of the population mean.

Part c

Now we have everything in order to replace into formula (1):

$7.14-1.83\frac{0.87}{\sqrt{10}}=6.637$

$7.14+1.83\frac{0.87}{\sqrt{10}}=7.643$

So on this case the 90% confidence interval would be given by (6.636;7.643)

Part d

Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

Part e

If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

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