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Volume/485566: Suppose that a box has a volume of 100 centimeters and that the areaof the bottom of the box is 50 square centimeters. Find the height of the box. 1 solutions
Answer 332121 by solver91311(16868) on 2011-08-24 18:05:40 (Show Source):
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Triangles/485614: In a certain isosceles triangle, the base is seven less than twice the measure of one of the legs. The perimeter is 65 cm. Find the lengths of all three sides of the triangle. 1 solutions
Answer 332108 by solver91311(16868) on 2011-08-24 17:36:39 (Show Source):
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Equations/485570: , where x= -1
I have tried and tried, but keep getting a solution of 1 or more exactly (3/3). The back of my textbook gives an answer of (3/2). I just don't see how. 1 solutions
Answer 332084 by solver91311(16868) on 2011-08-24 16:51:37 (Show Source):
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Probability-and-statistics/485450: Three screws are drawn at random from a lot of 200 screws, 10 of which are defective. Find the probability that the screws drawn will be nondefective in drawing (a) with replacement, (b) without replacement. 1 solutions
Answer 332047 by solver91311(16868) on 2011-08-24 15:01:44 (Show Source):
You can put this solution on YOUR website!
a) with replacement. The probability on each draw is constant, so use the Binomial distribution.
The probability of successes in trials where is the probability of success on any given trial is given by:
Where is the number of combinations of things taken at a time and is calculated by
You would be looking for:
Noting that pick 0 (and pick for that matter) equals 1 for all positive integers . Also for all real , so the calculation reduces to:
You can do your own arithmetic.
b)
The probabilities for each draw vary because you don't replace the drawn item.
The probability of a non-defective item on the first draw where there are 10 out of 200 defective is
Assuming that a good item was drawn on the first draw, the new probability for the second draw becomes . You decreased the total number of items by one, so the denominator is decremented, but since you did not select a defective item, the number of non-defective items remains at a constant 190. (If you had selected a defective item on the first draw, game over -- we are calculating the probability of drawing non-defectives.)
Again, assuming a non-defective on draw 2, decrement the denominator and keep the numerator constant.
Then multiply the three probabilities:
Again, you are on your own with the arithmetic.
John

My calculator said it, I believe it, that settles it
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Age_Word_Problems/485425: I am currently 60 years old. My grand daughter is 12. At some time I will be three times her age. Write an equation which models how old each of us will be. 1 solutions
Answer 332012 by solver91311(16868) on 2011-08-24 13:02:25 (Show Source):
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Probability-and-statistics/485391: Respected Sir / Mam ,
Please help me to solve this question. I will be very grateful for your help .
My question is
If the overall percentage of success in the exam is 60, what is the probability that out of a group of 4 students, at least one has passed?
As the answer given is : 0.9744
Please provide me the steps for this question.
Thank you
1 solutions
Answer 331988 by solver91311(16868) on 2011-08-24 12:06:59 (Show Source):
You can put this solution on YOUR website!
Like many things in this life there is a hard way and an easy way to answer this question.
If 4 students take the test, then there are 5 situations that could possibly arise:
None pass, 1 passes, 2 pass, 3 pass, or all 4 pass.
Note for future reference that since these five outcomes are inclusive of all possible outcomes, the sum of the probabilities of each must be 1, because it is a certainty that one of the outcomes will occur.
Now, your question asks for the probability that at least one student passes. Looking at it in a straight-forward way, that would be the sum of the probabilities that exactly 1 passes plus exactly 2 pass plus...and so on.
Since "Pass/Fail" is an either/or outcome, and the probability of success for any given instance of one student taking one test is given as 60%, we know that the probability should be calculated using the binomial distribution.
The probability of successes in trials where is the probability of success on any given trial is given by:
Where is the number of combinations of things taken at a time and is calculated by
But you would need to perform this calculation four times and then sum the results. First you would need to calculate:
To get the probability for exactly 1, and then you would have to do exactly 2, and so on. In summary:
But fortunately, there is a much simpler way. From the point of view of your question, there are only two possible outcomes. Either 1 or more pass, or nobody passes. So if you take the probability that nobody passes and subtract that from 1, you get the probability that at least 1 passes.
Since we know that pick 0 is 1 for all positive integers and that for all real numbers ,
The above reduces to:
The arithmetic is yours to do.
John

My calculator said it, I believe it, that settles it
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Travel_Word_Problems/485107: a long distance runner has an average speed of 5 m/s during a race. how far does the runner travel in 25 minutes? express the answer in m, km, and miles.
and lastly,
A rock is dropped off the side of a bridge and hits the water below at 2.5 s later. what is the rocks velocity when it hits the water, the average velocity as it fell and the height of the bridge above the water. 1 solutions
Answer 331833 by solver91311(16868) on 2011-08-23 19:16:25 (Show Source):
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Mixture_Word_Problems/485051: 7 people meet and shake hands with one another, how many handshakes occurred?
using inductive reasoning, wrte a formula for the number of handshakes if the number of people is n.
the Fibonacci sequence consists of the pattern 1,1,2,3,5,8,13,...., what is the ninth term in the pattern?
look at the successive ratios of ne term to the next, make a conjecture?
list the first eight terms of the sequence formed by finding the differences of sucessive terms in the fibonacci sequence? 1 solutions
Answer 331811 by solver91311(16868) on 2011-08-23 17:52:45 (Show Source):
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sets-and-operations/484526: I'm having a problem with this set. The question is..
Prior to the 7:15 show at the local movie theater, 70 people visited the concession stand, Listed below is what they ordered
42 ordered popcorn
18 ordered candy
30 ordered a soda
10 ordered popcorn and candy
8 ordered soda and candy
12 ordered popcorn and soda
5 ordered popcorn, candy and soda
1. How many people ordered something other than popcorn, candy or soda?
2. How many people ordered popcorn and a soda but not candy?
3. How many people ordered candy or a soda but not popcorn?
4. How many people ordered popcorn or candy? 1 solutions
Answer 331533 by solver91311(16868) on 2011-08-22 18:05:20 (Show Source):
You can put this solution on YOUR website!
Draw a large rectangle. Inside of the rectangle draw three circles that partially overlap.
Label the circles P, C, and S.
In the very center where the three circles overlap, that is to say the one region on your diagram where you are inside of all three circles, put the number 5 representing the 5 people who ordered P, C, AND S.
The region where P and C overlap needs to contain the number of people who ordered ONLY P and C. The number 10 given for people who ordered P and C also includes those who ordered P, C, and S. So from the given 10 subtract those that ordered all three, that is 10 minus 5 = 5. And you put 5 into the P and C only region. Similarly, since 12 minus 5 equals 7, you put 7 in the P & S only region. Then, since the number who ordered P includes those who ordered P & C, those who ordered P & S, AND those who ordered all three, add up 5 plus 7 plus 5 = 17, then subtract 17 from the 42 we are told ordered P to arrive at 25 who ordered ONLY P. Use similar logic to fill in the rest of your diagram.
Next add all of the numbers that you have entered in your diagram. Subtract that result from 70, the number of people surveyed, to get the number of people who did not order P, S, OR C. Several of us only wanted ice cream. Write that number in the rectangle outside of all of the circles.
You should be able to answer all of the questions posed just from the data in your diagram.
John

My calculator said it, I believe it, that settles it
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Probability-and-statistics/484529: Prevalence of Alzheimer's disease (cases per 100 population)
Age group; males; females
65-69; 1.6; 0.0
70-74; 0.0; 2.2
75-79; 4.9; 2.3
80-84; 8.6; 7.8
85+; 35.0; 27.9
suppose an unrelated 77 year old man, 76 year old woman, and 82 year old woman are selected from a community. what is the probability that all three of these individuals have Alzheimer's Disease. 1 solutions
Answer 331529 by solver91311(16868) on 2011-08-22 17:51:42 (Show Source):
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Probability-and-statistics/484530: Prevalence of Alzheimer's disease (cases per 100 population)
Age group; males; females
65-69; 1.6; 0.0
70-74; 0.0; 2.2
75-79; 4.9; 2.3
80-84; 8.6; 7.8
85+; 35.0; 27.9
Suppose an unrelated 77 year old man, 76 year old woman, and 82 year old woman are selected from a community. What is the probability that at least one of the three people has Alzheimer's disease? 1 solutions
Answer 331528 by solver91311(16868) on 2011-08-22 17:51:10 (Show Source):
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