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Nonamiya [84]
1 year ago
7

Music Company A charges a $15 membership fee and $0.75 each to download a song. Music Company B charges $0.90 each to download a

song with no membership fee.
a. For what number of songs will both companies charge the same amount? Justify your answer using mathematics.
songs
You anticipate that you will download 150 songs. Which company should you choose if you want to spend the least amount of money? Justify your answer using mathematics.

Mathematics
1 answer:
Contact [7]1 year ago
3 0

The number of songs that both companies will charge the same amount is 100. And to pay the least amount of money we need to choose Company B.

<h3>Linear Equation: </h3>

The mathematical expressions that we used to find unknown numbers are called Linear Equations. These can be formed by variables and constant terms according to given conditions. Here we use variables like x, y, z, etc to represent unknown variables.

 

Here we have,

Music company A charges a $ 15 membership fee and a $ 0.75 for each downloaded song

Music company B charges $ 0.90 for each downloaded song without the membership fee

Given question

a. For what number of songs will both companies charge the same amount?  

Let for x songs both companies will charge the same amount

Cost to download x songs from Company A

=> 15 + 0.75x

Cost to download x songs from Company B

=>  0.90x

As we assumed both costs are the same

=> 15 + 0.75x = 0.90x

=> 0.90x - 0.75 = 15

=> 0.15x = 15

-> x = 100

Therefore,

The number of songs that both companies will charge the same amount is 100.

b) You anticipate that you will download 150 songs. Which company should you choose if you want to spend the least amount of money?  

Cost of 150 that charges company A =  15 + 150(0.75)  

= 15 + 150(0.75)  = 15 + 112.5 = 127.5  

Therefore,

Company A will charge $ 127.5 for 150 songs

Cost of 150 songs that charges company B = 150(0.90)

= $ 135  

Company B will charge $ 135 for 150 songs

Therefore,

Company B will charge $ 135 for 150 songs

Thus,

To pay the least amount of money we need to choose Company B.

Therefore,

The number of songs that both companies will charge the same amount is 100. And to pay the least amount of money we need to choose Company B.

Learn more about Linear Equations at

brainly.com/question/11897796

#SPJ4

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A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards. (a) If you have at least one ace, what is the probab
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Answer:

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Step-by-step explanation:

We can answer both questions using conditional probability.

(a) We need to calculate the probability of obtaining two aces given that you obtained at least one. Let's call <em>A</em> the random variable that determines how many Aces you have. A is a discrete variable that can take any integer value from 0 to 4. We need to calculate

P(A \geq 2 | A \geq 1) = P(A\geq 2 \cap A \geq 1) / P(A \geq 1)

Since having 2 or more aces implies having at least one, the event A \geq 2 \cap A \geq 1 is equal to the event A \geq 2. Therefore, we can rewrite the previous expression as follows

P(A \geq 2) / P(A \geq 1)

We can calculate each of the probabilities by substracting from one the probability of its complementary event, which  are easier to compute

P(A \geq 2) = 1 - P((A \geq 2)^c) = 1 - P((A = 0) \bigsqcup (A = 1)) = 1 - P(A = 0) - P (A = 1)

P (A \geq 1) = 1 - P ((A \geq 1)^c) = 1 - P(A = 0)

We have now to calculate P(A = 0) and P(A = 1).

For the event A = 0, we have to pick 13 cards and obtain no ace at all. Since there are 4 aces on the deck, we need to pick 13 cards from a specific group of 48. The total of favourable cases is equivalent to the ammount of subsets of 13 elements of a set of 48, in other words it is 48 \choose 13. The total of cases is 52 \choose 13. We obtain

P(A = 0) = {48 \choose 13}/{52 \choose 13} = \frac{48! * 39!}{52!*35!} \simeq 0.303  

For the event A = 1, we pick an Ace first, then we pick 12 cards that are no aces. Since we can pick from 4 aces, that would multiply the favourable cases by 4, so we conclude

P(A=1) = 4*{48 \choose 12}/{52 \choose 13} = \frac{4*13*48! * 39!}{52!*36!} \simeq 0.438      

Hence,  

1 - P(A = 1)-P(A=0) /1-P(A=1) = 1 - 0.438 - 0.303/1-0.303 = 0.371

We conclude that the probability of having two aces provided we have one is 0.371

b) For this problem, since we are guaranteed to obtain the ace of spades, we can concentrate on the other 12 cards instead. Those 12 cards have to contain at least one ace (other that the ace of spades).

We can interpret this problem as if we would have removed the ace of spades from the deck and we are dealt 12 cards instead of 13. We need at least one of the 3 remaining aces. We will use the random variable B defined by the amount of aces we have other that the ace of spades. We have to calculate the probability of B being greater or equal than 1. In order to calculate that we can compute the probability of the <em>complementary set</em> and substract that number from 1.

P(B \geq 1) = 1-P(B=0)

In order to calculate P(B=0), we consider the number of favourable cases in which we dont have aces. That number is equal to the amount of subsets of 12 elements from a set with 48 (the deck without aces). Then, the amount of favourable cases is 48 \choose 12. Without the ace of spades, we have 51 cards on the deck, therefore

P(B = 0) = {48 \choose 12} / {51 \choose 12} = \frac{48!*39!}{51!*36!} = 0.438

We can conclude

P(B \geq 1) = 1- 0.438 = 0.561

The probability to obtain at least 2 aces if we have the ace of spades is 0.561

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