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Scrat [10]
3 years ago
15

4. Suppose that you have an enormous grapefruit that is 92% water (by weight). The grapefruit weights 100 pounds. If the water c

ontent of the grapefruit evaporates until it is 90% water (by weight), then approximately how much does the grapefruit now weigh?
Mathematics
1 answer:
Anton [14]3 years ago
8 0

Answer:

80 pounds.

Step-by-step explanation:

Let W₁ be the initial weight of water in the grapefruit, W₂ be the weight of water in the grapefruit after evaporation, and M be the total weight of the grapefruit after evaporation. Assuming that, originally, the grapefruit weighed 100 pounds with a 92% water content and the weight of the "non-water" part of the fruit (100 - W₁) does not change. The final weight of the grapefruit (M) is given by:

W_1=0.92*100=92\\W_2=0.90*M\\W_2+(100-W_1) = M\\W_2=M-8\\(1-0.9)M=8\\M=80

The grapefruit now weighs 80 pounds.

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According to the National Bridge Inspection Standard (NBIS), public bridges over 20 feet in length must be inspected and rated e
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Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that p = 0.09

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Either less than 4 have a rating of 4 or below, or at least 4 does. The sum of the probabilities of these events is 1.

So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225

P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827

P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082

P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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