I need help on this problem!!

Mathematics

1 answer:

Svetllana [295]2 years ago

4 0

$3.98*10^{6}(people)*\frac{1 mi^{2}}{1000 people} = 3.98*10^{6}* \frac{1 mi^{2}}{10^{3}} = 3.98*10^{6-3} mi^{2}=

=3.98*10^{3} mi^{2}$

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denis-greek [22]

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8 0

2 years ago

Marvin lives in Stormwind City and works as an engineer in the city of Ironforge. In the morning, he has 333 transportation opti

sukhopar [10]

The event "Atleast once" is the complement of event "None".

So, the probability that Marvin teleports atleast once per day will the compliment of probability that he does not teleports during the day. Therefore, first we need to find the probability that Marvin does not teleports during the day.

At Morning, the probability that Marvin does not teleport = 2/3

Likewise, the probability tha Marvin does not teleport during evening is also 2/3.

Since the two events are independent i.e. his choice during morning is not affecting his choice during the evening, the probability that he does not teleports during the day will be the product of both individual probabilities.

So, the probability that Marvin does not teleport during the day = $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$