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Savatey [412]
2 years ago
5

The disc containing the only copy of your homework just got corrupted, and the disk got mixed up with two other corrupted discs

that were lying around. It is equally likely that any of the three discs holds the corrupted remains of your homework. Your computer expert friend offers to have a look at one of the discs, and you know from past experience that his probability of finding your homework from a disc is 0. 35, given that the homework is in there. Given that he searches on disc 1 but cannot find your homework, what is the probability that your homework is on disc i, for i = 1, 2, 3?.
SAT
2 answers:
AleksAgata [21]2 years ago
6 0

The probability that your homework is on disc i = 12.25 %

Given data :

Probability of finding homework on a disc ( p ) = 0.35

number of disks ( i ) = 1,2,3

<h3>Probability calculation </h3>

To determine the probability of disc i containing your homework we will apply the calculation below

 

P( Disc i ) = p * p

where : p = probability of finding homework on a disc  

Therefore : P ( Disc i ) = 0.35 * 0.35

                                    = 0.1225

                                    = 12.25%

Hence we can conclude that,the probability of your homework been on disc i is : 12.25%.

Learn more about probability : brainly.com/question/25870256

horrorfan [7]2 years ago
5 0

The probability that your homework is on disc i is 12.25%.

<h3><u>Probabilities </u></h3>

Given that the disc containing the only copy of your homework just got corrupted, and the disk got mixed up with two other corrupted discs that were lying around, and it is equally likely that any of the three discs holds the corrupted remains of your homework, and your computer expert friend offers to have a look at one of the discs, and you know from past experience that his probability of finding your homework from a disc is 0. 35, given that the homework is in there, given that he searches on disc 1 but cannot find your homework, to determine what is the probability that your homework is on disc i, for i = 1, 2, 3, the following calculation must be performed:

  • Disc i = 0.35 x 0.35
  • Disc i = 0.35^2
  • Disc i = 0.1225
  • 0.1225 x 100 = 12.25

Therefore, the probability that your homework is on disc i is 12.25%.

Learn more about probability in brainly.com/question/26444510

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PilotLPTM [1.2K]

Answer:

1990 and 2010

y_l_w(15)=315\hspace{3}thousands\\y_l_c(15)=315\hspace{3}thousands\\y_l_w(35)=715\hspace{3}thousands\\y_l_c(35)=715\hspace{3}thousands

Explanation:

Let:

y_l_w=Population\hspace{3}of\hspace{3}Lewiston\\y_l_c=Population\hspace{3}of\hspace{3}Lockport

We need to know, in what year(s) the villages had the same population, mathematically this is:

y_l_w=y_l_c

So:

x^2-30x+540=20x+15\\\\Subtract\hspace{3}20x\hspace{3}from\hspace{3}both\hspace{3}sides\\\\x^2-50x+540=15\\\\Subtract\hspace{3}15\hspace{3}from\hspace{3}both\hspace{3}sides\\\\x^2-50x+525=0

Solving for x:

Factoring

(x-15)(x-35)=0

Hence:

x=15\\\\or\\\\x=35

Therefore the year(s) which the village had the same population are:

1975+15=1990\\\\and\\\\1975+35=2010

In order to find the population of both cities during the year(s) of equal population, just evalue the equations at x=15 and x=35:

y_l_w(15)=315\hspace{3}thousands\\y_l_c(15)=315\hspace{3}thousands\\y_l_w(35)=715\hspace{3}thousands\\y_l_c(35)=715\hspace{3}thousands

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