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LiRa [457]
4 years ago
10

Sorry about that, thank you for pointing that out.

Mathematics
2 answers:
Naily [24]4 years ago
7 0

Make two equations and solve using the substitution method.

8m + 4g = 49

3m + 2g = 21

where m = movies and g = video games

____________________________________________________________

Solve for g in the first equation.

8m + 4g = 49

Subtract 8m from both sides.

4g = 49 - 8m

Divide both sides by 4.

g = 49/4 - 2m

____________________________________________________________

Plug g into the second equation.

3m + 2(49/4 - 2m) = 21

Distribute 2 inside the parentheses.

3m + 24.5 - 4m = 21

Combine like terms.

-m + 24.5 = 21

Subtract 24.5 from both sides.

-m = -3.5

Divide both sides by -1.

m = 3.5

____________________________________________________________

Plug m into the first equation.

8(3.5) + 4g = 49

Distribute 8 inside the parentheses.

28 + 4g = 49

Subtract 28 from both sides.

4g = 21

Divide both sides by 4.

g = 5.25

____________________________________________________________

Rental cost for each movie: $ 3.50

Rental cost for each video game: $5.25

Marina CMI [18]4 years ago
4 0

The rental cost for each movie would be $3.50, and the rental cost for each video game would be $5.25.

If movies are 'm', and video games are 'v', we can write a set of simultaneous equations to solve:

8m + 4v = 49

3m + 2v = 21

We can solve using elimination, and multiply the second equation by 2 to get 6m + 4v = 42, and then subtract:

8m + 4v = 49

- 6m + 4v = 42

= 2m = 7

÷ 2

m = 3.5

Now we can substitute in 'm' as 3.5 to find 'v':

(3 × 3.5) + 2v = 21

10.5 + 2v = 21

- 10.5

2v = 10.5

÷ 2

v = 5.25

I hope this helps!

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oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

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

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

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

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

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

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

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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