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zhenek [66]
3 years ago
7

Mark transferred songs from his computer onto his portable music player. He transferred 2 6/7 songs in 1 2/3 minutes. How many s

ongs did he transfer per minute?
Mathematics
2 answers:
aleksklad [387]3 years ago
7 0

Answer:

1\frac{5}{7} songs/ minute

Step-by-step explanation:

Mark transferred songs from his computer onto his portable music player.

Since he transferred 2\frac{6}{7} songs in 1\frac{2}{3} minute

2\frac{6}{7} songs = \frac{20}{7} songs

1\frac{2}{3} minutes = \frac{5}{3} minutes

Therefore, he transfer songs per minute = \frac{\frac{20}{7} }{\frac{5}{3} }

                                                       = \frac{20}{7} × \frac{3}{5}

                                                      =  \frac{4}{7} ×  \frac{3}{1}

                                                      =  \frac{12}{7}

                                                      =  1\frac{5}{7} songs/ minute

Mark transferred  1\frac{5}{7} songs per minute.

dimulka [17.4K]3 years ago
4 0
This is basically unit rate. So we make them into mixed numbers. 2 6/7 = 12/7. 1 2/3 = 5/3. Then we have to divide them. 12/7 divided by 5/3= 12/7 times 3/5 because u have to flip the second one. Keep, Change, and Flip is my motto. Then u will get 36/35. Simplified it will be 1 1/35. I also never said it was right, so if u get it wrong, because of me I'm sorry I tried.
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Answer:

(a) The value of P (M | V) is 0.30.

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Step-by-step explanation:

It is provided that,

<em>V</em> = a student has a Visa card

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N = 100, n (<em>V</em>) = 40, n (<em>M</em>) = 32 and n (<em>V</em> ∩ <em>M</em>) = 12.

The probability of a student having visa card is:

P(V) = \frac{n(V)}{N}= \frac{40}{100}=0.40

The probability of a student having master card is:

P(M) = \frac{n(M)}{N}= \frac{32}{100}=0.32

The probability of a student having  visa card and a master card is:

P(V\cap M) = \frac{n(V\cap M)}{N}= \frac{12}{100}=0.12

The conditional probability of an event, say A, given that another event, say B, has already occurred is,

P(A|B)=\frac{P(A\cap B)}{P(B)}

(a)

Compute the probability that a student has a master card given that he/she has a visa card also, i.e. P (M | V) as follows:

P(M|V)=\frac{P(V\cap M)}{P(V)} =\frac{0.12}{0.40}=0.30

Thus, the value of P (M | V) is 0.30.

(b)

Compute the probability that a student does not have a master card given that he/she has a visa card also, i.e. P (M^{c} | V) as follows:

P (M^{c} | V)=1-P(M|V)=1-0.30=0.70

Thus, the value of P (M^{c} | V) is 0.70.

(c)

Compute the probability that a student has a visa card given that he/she has a master card also, i.e. P (V | M) as follows:

P(V|M)=\frac{P(V\cap M)}{P(M)} =\frac{0.12}{0.32}=0.375

Thus, the value of P (V | M) is 0.375.

(d)

Compute the probability that a student does not have a visa card given that he/she has a master card also, i.e. P(V^{c}|M) as follows:

P(V^{c}|M)=1-P(V|M)=1-0.375=0.625

Thus, the value of P(V^{c}|M) is 0.625.

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