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Serga [27]
2 years ago
14

6 divided by 603 is?

Mathematics
2 answers:
s2008m [1.1K]2 years ago
7 0

Note that both numbers are multiple of 3: you can simplify the ratio as follows

\dfrac{6}{603} = \dfrac{3\cdot 2}{3\cdot 201} = \dfrac{2}{201}

The ratio can't be further simplified because 2 and 201 are coprime. So, the only thing you could do is write down the decimal expansion of the number.

For the record, its period is 33 digits long, and it is

0.009950248756218905472636815920398


LekaFEV [45]2 years ago
5 0

Answer:

0.00995025

Step-by-step explanation:

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Step-by-step explanation:

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A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A st
Anon25 [30]

Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = an US adult who does not uses social media.

<em>Y</em> = an US adult between the ages 18 and 29.

<em>Z</em> = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

P (Z) = 0.78

P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (<em>X'</em><em>)</em>) = 1 - P (US adult so not use social media)

                                                   =1-P(X)\\=1-0.35\\=0.65

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (<em>Y</em>)) = 1 - P (Adults 30 or above)

                                            =1-P(Z)\\=1-0.78\\=0.22

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

                =0.22+0.65-0.672\\=0.198

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

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