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katrin2010 [14]
3 years ago
6

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.3 in. long with 9.0 lb hung f

rom the balance. How much work is done in stretching it from 6.0 in. to a length of 12.2 in.
Mathematics
1 answer:
Aleks04 [339]3 years ago
5 0

Answer:

74.958 lb-in

Step-by-step explanation:

We are given that

Natural length of spring=l=6 in

After stretching, the length of  spring=8.3 in

Weight=W=9 lb

Extension in the sprig=8.3-6=2.3 in

By hook's law

F==W=kx

Using the formula

9=k(2.3)

k=\frac{9}{2.3}=3.9 lb/in

Extension in the spring when the length of spring after stretching to 12.2 in=12.2-6=6.2 in

Work done, W=\int_{0}^{6.2}3.9xdx

W=[\frac{3.9x^2}{2}]^{6.2}_{0}

W=3.9\times \frac{(6.2)^2}{2}

W=74.958 lb-in

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In Sweden, 10 people are struck by lightning every year, on average. Answer thefollowing questions about a given year in Sweden
emmainna [20.7K]

Answer:

a) 19.85% probability that a total of two people are struck by lightning during first four months of the year.

b) 22.68% probability that the year has 5 good and 7 bad months

Step-by-step explanation:

We are going to use the Poisson distribution and the binomial distribition to solve this question.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

Binomial 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.

a.Find the probability that a total of two people are struck by lightning during first four months of the year.

10 people during a year(12 months).

In 4 months, the mean is \mu = \frac{10*4}{12} = 3.33

This is P(X = 2).

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 2) = \frac{e^{-3.33}*(3.33)^{2}}{(2)!} = 0.1985

19.85% probability that a total of two people are struck by lightning during first four months of the year.

b.Say that a month is good is no one is struck by lightning, and bad otherwise. Find the probability that the year has 5 good and 7 bad months.

Probability that a month is good.

P(X = 0), Poisson

The mean is \mu = \frac{10*1}{12} = 0.8333

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.8333}*(0.8333)^{0}}{(0)!} = 0.4346

Find the probability that the year has 5 good and 7 bad months.

Now we use the binomial distribution, we want P(X = 5) when n = 12, p = 0.4346. So

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

P(X = 5) = C_{12,5}.(0.4346)^{5}.(0.5654)^{7} = 0.2268

22.68% probability that the year has 5 good and 7 bad months

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