New!
Get regular updates about newly solved problems
via algebra.com's RSS system.
Recent problems solved by 'edjones'
edjones answered: 7568 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589, >>Next
Angles/468080: Which of these angles are complementary?A. 45o and 45o B. 63o and 117o C. 45o and 135o D. 90o and 45o 1 solutions
Answer 321161 by edjones(7569) on 2011-07-07 15:29:58 (Show Source):
|
Equations/468087: in a certain survey of 1,004 adults, 83% reported having more than one television at home. Find the actual number of respondents corresponding to the given percentage? 1 solutions
Answer 321157 by edjones(7569) on 2011-07-07 15:17:48 (Show Source):
|
Probability-and-statistics/468063: I would highly appreciate assistance from a tutor on how to do these three difficult homework problems that are due tonight. I don't know if I am solving them correctly. Please show me if I am mislead.
4. (1 point) Suppose in a village 50% of the people farm, 60% of the people fish, and 70% of the people either farm or fish. What is the percentage of people in the village who both farm and fish?
.5+.6= 1.1 - .7 = .4?
5. (1 point) Suppose in a village 50% of the people farm, 60% of the people fish, and 70% of the people either farm or fish. If a randomly chosen person farms, what is the probability that this person also fishes?
.5 x .6= .3?
6. (1 point) Suppose 48% of a village’s population is male. 30% of the men smoke, 20% of the women smoke. If a randomly chosen person is found to be a smoker, what is the probability that this person is a man? Round answer to two places after the decimal point.
.48 x .3 = .144= .14? 1 solutions
Answer 321146 by edjones(7569) on 2011-07-07 14:31:11 (Show Source):
You can put this solution on YOUR website!This is very easy once you understand Venn diagrams.
Let x=the intersection of farm and fish: .5-x+.6-x+x=.7 the union of farm and fish.
4)1.1-x=.7
x=.4 answer
farm only = .1
fish only = .2
.
5).4/.5=.8
.
6)
.48*.3=.144 male smokers
.52*.2=.104 female smokers
--------------add
======>.248
.
.144/.248 = .58 probability the this person is a man.
.
Ed
|
Probability-and-statistics/466654: I have worked this one out but I am not sure if the answer is correct. I only get one shot so any assistance helps.
A fair coin is flipped 20 times and the sequence of heads and tails is observed. Find the probability that the sequence consists of AT LEAST two heads. Answer in fraction reduced to the lowest terms.
i said:
1048555/1048576
Refer to the given graph. Assume one can move only down or to the right. If a route starts at A and ends at C, what is the probability that the route passes through B? Answer in fraction reduced to the lowest terms.
a
[][][]b[][]
[][][][][]
[][][][][]
[][][][][]
[][][][][]c
I said c(4,1) but that isn't what it's asking for. Do i need to multiply C(4,1) by c(10,5) to find the probability of passing through b? 1 solutions
Answer 320216 by edjones(7569) on 2011-07-03 22:02:04 (Show Source):
|
Probability-and-statistics/465878: What is the probability that 6 rolls of a fair die will show four exactly 2 times?
According to a college survey 22% of all students work fulltime. Find the mean for the random variable x the number of students who work full time in the sample is 16. 1 solutions
Answer 319274 by edjones(7569) on 2011-06-29 23:37:12 (Show Source):
|
Finance/465872: I really need some help here getting started...I do not know what I am suppose to do and there are many parts to this but a,b are the key questions I need help with because the other questions work off this.
a. Suppose a market research company finds that at a price of p = $10, they would sell x = 24 tiles each month. If they lower the price to p = $5, then more people would purchase the tile, and they can expect to sell x = 44 tiles in a month’s time. Find the equation of the line for the demand equation. Write your answer in the form p = mx + b. Hint: Write an equation using two points in the form (x,p) and remember that order matters!
b. Substitute the result you found from part a. into the equation R = xp to find the revenue equation. Provide your answer in simplified form. 1 solutions
Answer 319273 by edjones(7569) on 2011-06-29 23:26:57 (Show Source):
You can put this solution on YOUR website!(24,10) & (44,5)
.
(p-p1)/(x-x1)=m
m=(5-10)/(44-24)
=-5/20
=-1/4=-.25 slope
.
(p-p1)=m(x-x1)
p-10=-.25(x-24)
p-10=-.25x+6
p=-.25x+16
.
R=xp
=x(-.25x+16)
=-.25x^2+16x
.
Ed
|
Word_Problems_With_Coins/465770: A sum of money amounts to 9.20 pesos,is made up of 10Centavos and 25Centavos coins. If there are 8 more than coins in 10Centavos than 25Centavos. How many coins each kind are there?
1 solutions
Answer 319258 by edjones(7569) on 2011-06-29 21:49:35 (Show Source):
|
Permutations/465776: 1. How many garlands can be made from 6 marigolds and 2 roses?
2. How many ways can a necklace be formed from 2 red and 2 blue beads? 1 solutions
Answer 319254 by edjones(7569) on 2011-06-29 21:37:07 (Show Source):
|
Probability-and-statistics/465515: A sample of 400 racing cars showed that 80 cars cost over $700,000. What is the 99% confidence interval of the true proportion of cars costing over $700,000? 1 solutions
Answer 319016 by edjones(7569) on 2011-06-28 23:05:28 (Show Source):
You can put this solution on YOUR website! , q=0.8, n=400

(a,b)=  *  =.2+-(2.576*.02)=(.1485, .2515)
We can say with 99% confidence that the proportion of cars costing over $700,000 is in the interval (.1485, .2515).
.
Ed
|
Evaluation_Word_Problems/465513: For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015 (BusinessWeek, March 20, 2006). Assume the standard deviation is $3,650 and that debt amounts are normally distributed.
What is the probability that the debt for a randomly selected borrower with good credit is more than $18,000 (to 4 decimals)?
What is the probability that the debt for a randomly selected borrower with good credit is less than $10,000 (to 4 decimals)?
What is the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 (to 4 decimals)?
What is the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 (to 4 decimals)?
1 solutions
Answer 319010 by edjones(7569) on 2011-06-28 22:24:45 (Show Source):
You can put this solution on YOUR website!What is the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 (to 4 decimals)?
z=(12000-15015)/3650 = -.8260
z=(18000-15015)/3650 = .8178
The area under the normal curve between -.8260 and .8178 is .5889 the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000
.
Ed
|
Probability-and-statistics/465440: According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20, 2006). Assume the standard deviation is .7 hours and that the probability distribution is normal.
What is the probability that a randomly selected person sleeps more than 8 hours (to 4 decimals)?
What is the probability that a randomly selected person sleeps 6 hours or less (to 4 decimals)?
Doctors suggest getting between 7 and 9 hours of sleep each night. What percentage of the population gets this much sleep (to the nearest whole number)?
% 1 solutions
Answer 319007 by edjones(7569) on 2011-06-28 22:15:45 (Show Source):
You can put this solution on YOUR website!z=(8-6.8)/.7=1.714
Area under normal curve above 1.714 is .0433 the probability that a randomly selected person sleeps more than 8 hours.
.
Ed
|
Probability-and-statistics/465510: A study found that the average life expectancy is 53 years with a standard deviation of 7.5 years. If a person is selected at random, what is the probability that the person will die before the age of 65?
A. 94.52%
B. 82.89%
C. 94.95%
D. 88.49% 1 solutions
Answer 319004 by edjones(7569) on 2011-06-28 22:07:34 (Show Source):
|
Evaluation_Word_Problems/465438: For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015 (BusinessWeek, March 20, 2006). Assume the standard deviation is $3,380 and that debt amounts are normally distributed.
What is the probability that the debt for a randomly selected borrower with good credit is more than $18,000 (to 4 decimals)?
What is the probability that the debt for a randomly selected borrower with good credit is less than $10,000 (to 4 decimals)?
What is the probability that the debt for a randomly selected borrower with good credit is between $12,000 and $18,000 (to 4 decimals)?
What is the probability that the debt for a randomly selected borrower with good credit is no more than $14,000 (to 4 decimals)?
1 solutions
Answer 319003 by edjones(7569) on 2011-06-28 22:02:25 (Show Source):
You can put this solution on YOUR website!z=(18000-15015)/3380=.8831
area under normal curve above z=.8831 is .1886 the probability that the debt for a randomly selected borrower with good credit is more than $18,000.
.
Ed
|
Probability-and-statistics/465475: SAT verbal scores are normally distributed with a mean of 430 and a standard deviation of 120
If 1 of the student is randomly selected, find the probability that he or she obtained a score higher than 440.
1 solutions
Answer 318996 by edjones(7569) on 2011-06-28 21:45:27 (Show Source):
You can put this solution on YOUR website!z=(440-430)/120=.083
The area under the normal curve above z=.083 is .467 the probability that he or she obtained a score higher than 440.
.
Ed
|
Probability-and-statistics/465499: I'm stuck on this :/ some one please help
SAT verbal scores are normally distributed with a mean of 430 and a standard deviation of 120 If 1 of the student is randomly selected, find the probability that he or she obtained a score higher than 440.
1 solutions
Answer 318991 by edjones(7569) on 2011-06-28 21:26:14 (Show Source):
You can put this solution on YOUR website!z=(440-430)/120=.083
The area under the normal curve above z=.083 is .467 the probability that he or she obtained a score higher than 440.
.
Ed
|
Permutations/465104: Find the indicated probability. Express your answer as a simplified fraction unless otherwise noted.
The following table contains data from a study of two airlines which fly to Small Town, USA.
# on time # late
Poduck airlines 33 6
Upstate Airlines 43 5
If one of the 87 flights is randomly selected, find the probability that the flight selected arrived on time given that it was an Upstate Airlines flight 1 solutions
Answer 318733 by edjones(7569) on 2011-06-28 03:23:56 (Show Source):
|
|