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nadezda [96]
3 years ago
6

In reducing one's speed from 70 mph to 50 mph, how much of a percentage decrease in stopping distance is realized (to the neares

t percent)?

Mathematics
1 answer:
labwork [276]3 years ago
6 0

we are given

we are given relation between speed and distance

distance ( when speed =70mph) is

=328feet

so, d_1=328

distance ( when speed =50mph) is

=183feet

so, d_2=183

now, we can find percentage in decrease

=\frac{d_1-d_2}{d_1} \times 100

now, we can plug values

=\frac{328-183}{328} \times 100

=44%...........Answer

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- than 3,4 and 5 mean right triangle so 2,5 and 4 mean acute triangle sure or you can prove it in this way too
- than 3^2 +4^2 =5^2 => a right triangle so than 
2^2 +4^2 < 5^2 => an acute triangle 

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3 years ago
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

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

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
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