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irga5000 [103]
3 years ago
6

A music website charges x dollars for individual songs and y dollars for entire albums. Person A pays $25.92 to download 6 indiv

idual songs and 2 albums. Person B pays $33.93 to download 4 individual songs and 3 albums. Write a system of linear equations that represents this situation. How much does the website charge to download a song? an entire album?
Mathematics
1 answer:
kati45 [8]3 years ago
7 0
Amount charged for downloading individual songs = x dollars
Amount charged for downloading an entire album = y dollars
In respect to person A:
6x + 2y = 25.92
3x + y = 12.96
y = 12.96 - 3x
In respect to person B:
4x + 3y = 33.93
Putting the value of y from the first equation in the second, we get
4x + 3(12.96 - 3x) = 33.93
4x + 38.88 - 9x = 33.93
- 5x = - 4.95
x = 0.99 dollars
Putting the value of x in the first equation, we get
y = 12.96 - 3x
   = 12.96 - 2.97
   = 9.99 dollars

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48.41% probability that 17 or fewer of them are bilingual.

Step-by-step explanation:

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 50, p = \frac{466}{1320} = 0.3530

So

\mu = E(X) = np = 50*0.3530 = 17.65

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.3530*0.6470} = 3.38

Probability that 17 or fewer of them are bilingual.

Using continuity correction, this is P(X \leq 17 + 0.5) = P(X \leq 17.5), which is the pvalue of Z when X = 17.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{17.5 - 17.65}{3.38}

Z = -0.04

Z = -0.04 has a pvalue of 0.4841

48.41% probability that 17 or fewer of them are bilingual.

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