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2 years ago

13

A game increased in price by 1/2 . After the increase it was priced at £27. What was the original price of the game? (Hegarty Ma

ths)(SSSOOOO hard)(anyone know?)

2 answers:

jok3333 [9.3K]2 years ago

7 0

Answer:

£18

Step-by-step explanation:

Step-by-step explanation:

Let

x = original price of the game

Increase in price = 1/2

New price = £27

x + 1/2x = £27

2x+x/2 = 27

3/2x = 27

x = 27 ÷ 3/2

= 27 × 2/3

= 54 / 3

x = £18

Therefore, the original price of the game is £18

lukranit [14]2 years ago

4 0

Answer:

£18

Step-by-step explanation:

Let

x = original price of the game

Increase in price = 1/2

New price = £27

x + 1/2x = £27

2x+x/2 = 27

3/2x = 27

x = 27 ÷ 3/2

= 27 × 2/3

= 54 / 3

x = £18

Therefore, the original price of the game is £18

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Answer:

$320128-1.96\frac{50324}{\sqrt{529}}=315839.52$

$320128+1.96\frac{50324}{\sqrt{529}}=324416.48$

So on this case the 95% confidence interval would be given by (315839.52;324416.48)

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".

$\bar X=320128$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=50324$ represent the population standard deviation

n=529 represent the sample size

Solution to the problem

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

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

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$320128-1.96\frac{50324}{\sqrt{529}}=315839.52$

$320128+1.96\frac{50324}{\sqrt{529}}=324416.48$

So on this case the 95% confidence interval would be given by (315839.52;324416.48)

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