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2 years ago

during a flu epidemic,146 students out of 680 who attend Lincoln middle school were absent. what percent were absent?

Mathematics

1 answer:

ra1l [238]2 years ago

5 0

21% were absent, if you divide 146 by 680 and multiply by 100, then round to the nearest whole number, you get 21

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Help please!!!!!!!!!!!!!

Annette [7]

Answer:

Hm? Is there meant to be a file, if so I can’t see it. :((

Step-by-step explanation:

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2 years ago

Frank bought 2 CDs that were each the same price. Including sales tax, he paid a total of 24.20 Each CD had a tax of 0.60 What w

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Answer:

11.5

Step-by-step explanation:

Frank paid a total of 24.2 for the 2 CDs, so, for each CD, he paid the total value divided by two:

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Each CD had a tax of 0.6, so, to calculate the price of each CD before tax, we just subtract the price of each CD by the value of the tax for each CD:

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3 years ago

Solve: XC - XV. Show your answer as a Roman Numeral. Standard Roman Number Numeral 1 1 5 V 10 Х 50 L 100 С C 500 D 1,000 M OA) V

Keith_Richards [23]

Answer:

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Step-by-step explanation:

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2 years ago

Clara created the poster shown below:

lana66690 [7]

Answer:

$length=4\ cm\\width=6\ cm$

Step-by-step explanation:

Given the rectangular poster shown in the picture, you know that its dimensions (its width and its lenght) are:

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In order to calculate the dimensions of the rectangular at $\frac{1}{4}$ times its current size, you need to multiply the original lenght by $\frac{1}{4}$ and multiply the original width by $\frac{1}{4}$ .

Knowing this, you get:

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2 years ago

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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 = $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$

Probability that Marvin teleports atleast once during the day = 1 - Probability that Marvin does not teleport during the day.

Probability that Marvin teleports atleast once during the day = $1-\frac{4}{9}=\frac{5}{9}$

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2 years ago