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melamori03 [73]
3 years ago
12

Almost a billion people on the planet don't have access to clean drinking water." if there were 7.2×109 people in the world in 2

014, then approximately what percent of them lived without clean drinking water? round to the nearest tenth of a percent.
Mathematics
2 answers:
Ivahew [28]3 years ago
7 0
13.9%.

7.2 x 10^9 = 7,200,000,000.

Since 1 billion = 1,000,000,000 out of these 7,200,000,000 do not have access to clean water, we find the percent by taking:
1,000,000,000/7,200,000,000 = 10/72 = 13.9%
Leokris [45]3 years ago
5 0

Answer: 13.9%

Step-by-step explanation:

Given : Almost a billion people on the planet don't have access to clean drinking water."

We know that 1 billion = 1,000,000,000=10^9

There were 7.2\times10^9 people in the world in 2014.

Now, the percent of people lived without clean drinking water :-

\dfrac{\text{Number of people lived without clean water}}{\text{Total people in the world}}\times100\\\\=\dfrac{10^9}{7.2\times10^9}\times100\\\\=\dfrac{1}{7.2}\times100\\\\=0.138888888889\times100\\\\=13.8888888889\%\approx13.9\% \ \ \text{[Round to the nearest tenth of a percent]}

Hence, approximately 13.9% of them lived without clean drinking water.

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Answer in pic

Step-by-step explanation:

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3 years ago
Lunch at school consists of a sandwich, a vegetable, and a fruit. Each lunch combination is equally likely to be given to a stud
daser333 [38]

Answer:

a) see attachment

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c) The probability that Sol gets at least one of his favorite = 5/6

Step-by-step explanation:

a) A tree diagram is used to represent sample space.

We are told the lunch choices are sandwich, a vegetable and a fruit and each lunch combination are possible.

Find attached the tree diagram.

From the tree diagram, there are a total of 12 possible outcomes.

b) Sol's favorite lunch is a chicken sandwich, carrots, and a banana.

The probability that Sol gets his favorite lunch = Pr(chicken sandwich, carrots and a banana)

Pr(chicken sandwich, carrots, and a banana) = 1/12

Reason: From the tree diagram, the likelihood of having all three (chicken sandwich, carrots and a banana) occurs only once.

Pr(chicken sandwich, carrots and a banana) = [number of times(chicken sandwich, carrots and a banana) occurs]/(total number of possible outcomes)

Pr(chicken sandwich, carrots, and a banana) = 1/12

c) For Sol to get at least one of his favorite, we would have at least either chicken, carrot or banana in the possible outcome.

We have 10 of such possible outcome:

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The probability that Sol gets at least one of his favorite = (total number of at least one of Sol's favorite lunch)/(total possible outcome)

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The probability that Sol gets at least one of his favorite = 5/6

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