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Sonbull [250]
1 year ago
6

5. Find the selling Price of an item that originally cost a store $70 was marked up20% and then after no selling for 6 months wa

s discounted 10%?
Mathematics
1 answer:
zavuch27 [327]1 year ago
4 0

Answer:

$75.6

Explanation:

If the original cost was $70 and it was marked up 20%, the price after this is calculated as:

$70 + $70(0.2) = $70 + $14 = $84

Where $70(0.2) = $14 is the amount added to the original when it is marked up 20%.

Then, it was discounted 10%, we get that the selling price is

$84 - $84(0.1) = $84 - $8.4 = $75.6

Where $84(0.1) = $8.4 is the discount made on the initial price.

Therefore, the answer is $75.6

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For this case, we must find an expression equivalent to:

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Rewriting the previous expression we have:

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According to one of the properties of powers of the same base, we must put the same base and add the exponents:

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Answer:

\frac {3} {5x ^ 6 * y ^ 6}

Option B

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Answer:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a.2

Step-by-step explanation:

1) Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu=?,\sigma=?)  

We are interested on this probability

P(X>X+3\mu)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

And we can find the following probablity:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

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