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Recent problems solved by 'stanbon'
stanbon answered: 57351 problems
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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 , 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56970..56999 , 57000..57029 , 57030..57059 , 57060..57089 , 57090..57119 , 57120..57149 , 57150..57179 , 57180..57209 , 57210..57239 , 57240..57269 , 57270..57299 , 57300..57329 , 57330..57359, >>Next
Probability-and-statistics/347921: In a blood testing procedure, blood samples from 6 people are combined into one mixture. The mixture will only test negative if all the individual samples are negative. If the probability that an individual sample tests positive is 0.11, what is the probability that the mixture will test positive
1 solutions
Answer 248759 by stanbon(57424)   on 2010-09-25 15:23:12 (Show Source):
You can put this solution on YOUR website!In a blood testing procedure, blood samples from 6 people are combined into one mixture. The mixture will only test negative if all the individual samples are negative.
If the probability that an individual sample tests positive is 0.11, what is the probability that the mixture will test positive
----
Tests positive if one or more is positive.
----
P(tests negative) = 0.89^6 = 0.4950
---
P(tests positive) = 1-0.4950 = 0.5030
==================================
Cheers,
Stan H.
|
Probability-and-statistics/347933: In a large vase, there are 8 roses, 5 daisies, 12 lilies, and 9 orchids. If 4 flowers are selected at random, find the probability that at least 1 of the flowers is a rose.
1 solutions
Answer 248758 by stanbon(57424) on 2010-09-25 15:13:07 (Show Source):
You can put this solution on YOUR website!In a large vase, there are 8 roses, 5 daisies, 12 lilies, and 9 orchids. If 4 flowers are selected at random, find the probability that at least 1 of the flowers is a rose.
-------------
P(at least one rose in picking 4) = 1 - P(no rose in picking 4)
= 1 - [26C4/34C4]
---
= 1 - 0.3224
---
= 0.6776
==============
Cheers,
Stan H.
|
test/347897: Can you please help me subtract and simplify by collecting like radical terms, if possible?
a 3sqrt 7 - 2 3sqrt 7 =
Thanks
1 solutions
Answer 248746 by stanbon(57424) on 2010-09-25 11:21:48 (Show Source):
You can put this solution on YOUR website!(3sqrt(7)) - 2*(3sqrt(7))
-----
Factor:
(3sqrt(7)(1-2)
=(3sqrt(7))(-1)
---
= -3sqrt(7)
==================
Cheers,
Stan H.
==================
|
Exponents/347909: Please explain the difference between –122 and (–12)2
1 solutions
Answer 248744 by stanbon(57424) on 2010-09-25 11:18:00 (Show Source):
You can put this solution on YOUR website!Please explain the difference between –12^2 and (–12)^2
------------------------
-12^2
Note: You only squaring the 12.
= -(12*12) = -144
----
(-12)^2
Note: Here you are squaring (-12)
= (-12)(-12) = +144
================================
Cheers,
Stan H.
|
Exponential-and-logarithmic-functions/347900: A club sold fruitcakes for a fund raising project. The two-pound cakes sold for $5 each and the five-pound cakes sold for $11 each. The club sold 323 pounds of cakes and received $749. How many of each did they sell?
1 solutions
Answer 248734 by stanbon(57424) on 2010-09-25 10:33:27 (Show Source):
You can put this solution on YOUR website!A club sold fruitcakes for a fund raising project. The two-pound cakes sold for $5 each and the five-pound cakes sold for $11 each. The club sold 323 pounds of cakes and received $749. How many of each did they sell?
-------------------
Quantity Eq.::: 2t + 5f = 323 lbs
---
Value Eq: 5(t) + 11(f) = 749 dollars
----------------------------------------
Multiply thru 1st by 5
Multiply thru 2nd by 2
--------------------------
10t + 25f = 5*323
10t + 22f = 2*749
----------------------
Subtract 2nd from 1st and solve for "f":
3f = 117
f = 29 (# of 5-pound cakes)
---
Use 2t + 5f = 323 to solve for "t":
2t + 5*29 = 323
2t = 178
t = 89 (# of 2-pound cakes)
==============================
Cheers,
Stan H.
|
Exponential-and-logarithmic-functions/347898: Find the solution(s), if any, to the system
4x-6y=12
6x-9y=15
1 solutions
Answer 248732 by stanbon(57424) on 2010-09-25 10:21:41 (Show Source):
You can put this solution on YOUR website!4x-6y=12
6x-9y=15
--------------
Divide thru the 1st equation by 2.
Divide thru the 2nd equation by 3.
----
2x - 3y = 6
2x - 3y = 5
----
Subtract:
0 = 1
-----
That contradiction occurred because I assumed
the x's were the same and the y's were the
same in the two equations.
---
They are not.
The system is inconsistent.
The lines are parallel.
There is no solution for this system of equations.
==========================
Cheers,
Stan H.
=============
|
test/347896: Can you help me with adding and simplify by collecting like terms if possible, and assuming that all expressions under radicals represent nonnegative numbers.
sqrt 3a + 9 sqrt 27a^3
Thanks
1 solutions
Answer 248731 by stanbon(57424) on 2010-09-25 10:15:18 (Show Source):
You can put this solution on YOUR website!sqrt 3a + 9 sqrt 27a^3
---
sqrt(3a) + 9*sqrt(9a^2*3a)
---
sqrt(3a) + 9*3a*sqrt(3a)
---
Factor:
(sqrt(3a))(1 + 27a))
========================
Cheers,
Stan H.
|
Numbers_Word_Problems/347887: Planters Nuts Inc wants to make a 15lb of mixed nuts from peanuts that cost $4/lb and mixed nuts without peanuts that cost $7/lb. The cost of the final mixture should be $6/lb. How many lbs of peanuts should be used? Show equation..I am so stuck on this problem. I do not get this one at all
1 solutions
Answer 248730 by stanbon(57424) on 2010-09-25 09:59:57 (Show Source):
You can put this solution on YOUR website!Planters Nuts Inc wants to make a 15lb of mixed nuts from peanuts that cost $4/lb and mixed nuts without peanuts that cost $7/lb. The cost of the final mixture should be $6/lb. How many lbs of peanuts should be used? Show equation.
-------------------------
Let p be amount of peanuts
Let n be amount without peanuts.
---
Quantity Equation: p + n = 15 lb
Value Equation::: 4p + 7n = 6*15 dollars
--------------------------------------------
Multiply thru 1st by 7.
7p + 7n = 7*15
4p + 7n = 6*15
-------------------
Subtract the 2nd from the 1st and solve for "p":
3p = 15
p = 5 lb (amt. of peanuts used)
====================================
Cheers,
Stan H.
================
|
Linear-equations/347863: please answer sir,ma'am"mark contracted 15 skilled workers to finish cementing a certain walk of uniform width in 16 days.after 8 days of work,6 of them got sick and were not replaced until 4 days after.he wants the job to be on time .how many should he hire?"
thank you
1 solutions
Answer 248729 by stanbon(57424) on 2010-09-25 09:49:11 (Show Source):
You can put this solution on YOUR website!mark contracted 15 skilled workers to finish cementing a certain walk of uniform width in 16 days.
---
Total man-days is 15*16 = 240
-----------------
after 8 days of work 15*8 = 120 man-days of work are completed
----
With 6 men sick for 3 days, the 9 men complete 9*3 = 27 man-days
of work, leaving 93 man-days still to be done.
he wants the job to be on time .how many should he hire?"
---------------
There are 4 days remaining to do 223 man-days of work
93/4 = 24 1/4 men
---
He needs 10 additional men.
=============================
Cheers,
Stan H.
|
Polynomials-and-rational-expressions/347855: The problem says for me to solve but I don't even know how to begin. I know I need an LCM but what is it, 3x? Then what, do I divide both sides by the 3x? WHen I did that I came up with 6 = 21 + x
solve for x: 6-21=x
-15=x
Is this right?
2/x = 7/x - 1/3
1 solutions
Answer 248713 by stanbon(57424) on 2010-09-25 07:24:26 (Show Source):
You can put this solution on YOUR website!2/x = 7/x - 1/3
----
Multiply both sides by 3x to get rid of the denominators.
6 = 21 - x
x = 21-6
x = 15
=================
Cheers,
Stan H.
|
Probability-and-statistics/347754: There are 6 multiple-choice questions on an exam, each with four possible answers. Only one answer per question counts as the correct answer. If you guess on each question, what is the probability that you correctly answer all six questions?
1 solutions
Answer 248710 by stanbon(57424) on 2010-09-25 07:19:53 (Show Source):
You can put this solution on YOUR website!There are 6 multiple-choice questions on an exam, each with four possible answers. Only one answer per question counts as the correct answer. If you guess on each question, what is the probability that you correctly answer all six questions?
--------------
Ans: (1/4)^6
Cheers,
Stan H.
|
Probability-and-statistics/347840: Someone rolls a die and then picks cards from a standard deck equal to the number showing on the die.
a) What is the probability that the number of jacks in the hand equals 2?
b) Conditional on knowing the number of jacks in the hand equals 2, what is the conditional probability that the die showed 3?
1 solutions
Answer 248707 by stanbon(57424) on 2010-09-25 07:16:04 (Show Source):
You can put this solution on YOUR website!Someone rolls a die and then picks cards from a standard deck equal to the number showing on the die.
a) What is the probability that the number of jacks in the hand equals 2?
b) Conditional on knowing the number of jacks in the hand equals 2, what is the conditional probability that the die showed 3?
--------------------
Comment: How many cards are in the "hand"?
Cheers,
Stan H.
|
Probability-and-statistics/347865: Among the contestants in a competition of 35 women and 20 men. If 5 winners are randomly selected, what is the probability that they are all men? I do not understand where I am going wrong and thank you for the help.
1 solutions
Answer 248706 by stanbon(57424) on 2010-09-25 07:14:07 (Show Source):
You can put this solution on YOUR website!Among the contestants in a competition of 35 women and 20 men. If 5 winners are randomly selected, what is the probability that they are all men?
----------------
# of ways to pick 5 men: 20C5
# of ways to pick 5 randomly: 55C5
----
Probability of picking 5 men: 20C5/55C5 = 0.0045
-----------------------------------------------------
Cheers,
Stan H.
|
Probability-and-statistics/347866:
1 solutions
Answer 248705 by stanbon(57424) on 2010-09-25 07:11:13 (Show Source):
You can put this solution on YOUR website! A bin contains 77 light bulbs of which 12 are defective. If 5 light bulbs are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones. My original answer was .844 and it was wrong.
-----
Ans: (65/77)^5 = 0.4287
=======================
Cheers,
Stan H.
|
Probability-and-statistics/347868: This question involves tobacco use. Men that are non-smokers= 387, Ocassional smokers= 45, Regular Smoker= 90, Heavy Smoker= 37. Women that are non smoker=421, Ocassional smokers=46, Regular smokers =69, Heavy smokers =34.
If one of the 1129 people are randomly selected, find the probability that the person is a man or a heavy smoker. My original answer was .558 and I got it wrong, thank you so much for the help.
1 solutions
Answer 248704 by stanbon(57424) on 2010-09-25 07:07:09 (Show Source):
You can put this solution on YOUR website!This question involves tobacco use.
Men that are non-smokers= 387, Occasional smokers= 45, Regular Smoker= 90, Heavy Smoker= 37.
Women that are non smoker=421, Occasional smokers=46, Regular smokers =69, Heavy smokers =34.
--------------
If one of the 1129 people are randomly selected, find the probability that the person is a man or a heavy smoker. My original answer was .558 and I got it wrong, thank you so much for the help.
-----
P(man or hs) = P(man) + P(hs) -P(man and hs)
= (559/1129) + (71/1129) - (37/1129) = 0.5252
================================================
Cheers,
Stan H.
===============
|
Probability-and-statistics/347869: If two different people are randomly selected from the 875 subjects, find the probability that they are both heavy smokers. My original answer was .006040.
Men- Non-smokers =375, Light smokers= 34, Heavy Smoker= 35
Women Non-smokers =348,Light smokers =50,Heavy Smokers =33
1 solutions
Answer 248703 by stanbon(57424) on 2010-09-25 07:00:46 (Show Source):
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Miscellaneous_Word_Problems/347803: The probability of winning the lottery at Happytown is one in a million. Joe plays the lottery 500 times. What is the probability that he does eventually win?
1 solutions
Answer 248678 by stanbon(57424) on 2010-09-24 22:09:42 (Show Source):
You can put this solution on YOUR website!The probability of winning the lottery at Happytown is one in a million. Joe plays the lottery 500 times. What is the probability that he does eventually win?
Ans: 500/10^6 = 0.0005
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Cheers,
Stan H.
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Equations/347801: let x represent one number and let y represent the other number. the sum of two numbers is 6. if the second number is subtracted from the other, the result is 0. find the numbers
1 solutions
Answer 248675 by stanbon(57424) on 2010-09-24 22:07:10 (Show Source):
You can put this solution on YOUR website!let x represent one number and let y represent the other number. the sum of two numbers is 6. if the second number is subtracted from the other, the result is 0. find the numbers
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Equations:
x + y = 6
x - y = 0
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2x = 6
x = 3
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Since x+y = 6, y = 3
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Cheers,
Stan H.
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Age_Word_Problems/347796: abe and his younger sister are 3 years apart in age. if the sum of their ages will be 35 next year, what are their ages now?
1 solutions
Answer 248673 by stanbon(57424) on 2010-09-24 22:04:36 (Show Source):
You can put this solution on YOUR website!abe and his younger sister are 3 years apart in age. if the sum of their ages will be 35 next year, what are their ages now?
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Equations:
a-s = 3
a+s = 35
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Add and solve for "a":
2a = 38
a = 19 (abe's age)
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Since a+s = 35, s = 35-19 = 16 (his sister's age)
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Cheers,
Stan H.
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Probability-and-statistics/347795: A manufacture claims that 5% of the finished items coming off an assembly line each dat are defective. A total of 5,000 items are produced a day, and 3 are randomly selected and inspected.
a. If the manufacturer's claim is correct, what is the probability all 3 randomly selected items are defective?
b. If all 3 randomly selected items are defective, would you infer the manufacturer's claim is incorrect? why?
c. if the manufacturer's claim is correct, what is the probability none of the 3 items are defective?
d. If none of the 3 items are defective, would you infer that the manufacturer's claim is incorrect? why?
1 solutions
Answer 248669 by stanbon(57424) on 2010-09-24 22:00:09 (Show Source):
You can put this solution on YOUR website!A manufacture claims that 5% of the finished items coming off an assembly line each day are defective. A total of 5,000 items are produced a day, and 3 are randomly selected and inspected.
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a. If the manufacturer's claim is correct, what is the probability all 3 randomly selected items are defective?
(0.05)^3 = 0.000125
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b. If all 3 randomly selected items are defective, would you infer the manufacturer's claim is incorrect?
why?
No; the probability that would happen is extremely low.
The defect rate is probably higher.
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If the manufacturer's claim is correct, what is the probability none of the 3 items are defective?
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0.95^3 = 0.8574
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d. If none of the 3 items are defective, would you infer that the manufacturer's claim is incorrect?
why?
His claim has not really been tested. You would have no reason to
doubt it.
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Cheers,
Stan H.
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Expressions-with-variables/347792: Please help me solve the formula for the indicated variable and evaluate the written formula for the given values to the following:
Volume of a cone: V=Pie r squared h/3 Solve for h. Find h when V=25.12cm to the third and r=2cm. Use 3.14 for pie.
I am so totally lost. I would appreciate your help on this.
1 solutions
Answer 248667 by stanbon(57424) on 2010-09-24 21:50:33 (Show Source):
You can put this solution on YOUR website!Volume of a cone: V=Pie r squared h/3 Solve for h.
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Solve for "h":
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V = [(pi)r^2][h/3]
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h/3 = V/[(pi)r^2]
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h = (3V)/[(pi)r^2]
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Find h when V=25.12cm to the third and r=2 cm. Use 3.14 for pie.
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h = (3*25.12 cm^3)/[3.14*(2 cm)^2]
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h = (75.36)cm/[12.56]
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h = 6 cm
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Cheers,
Stan H.
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Miscellaneous_Word_Problems/347790: If a fair die rolled once comes up 6 you win $100. Otherwise, you lose $15. Would a rational person play this game? Justify your answer, using the definition of E(X).
1 solutions
Answer 248664 by stanbon(57424) on 2010-09-24 21:37:26 (Show Source):
You can put this solution on YOUR website!If a fair die rolled once comes up 6 you win $100. Otherwise, you lose $15. Would a rational person play this game? Justify your answer, using the definition of E(X).
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Probability of rolling a 6: 1/6
Probability of not rolling a 6: 5/6
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E(x) = (1/6)*100 + (5/6)(-15)
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E(x) = (100 - 75)/6 = 25/6 = $4.17
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The player can expect to win $4.17 each time he/she plays the game.
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Cheers,
Stan H.
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Equations/347769: find the inverse of: f(x)=3x+6
1 solutions
Answer 248659 by stanbon(57424) on 2010-09-24 21:32:53 (Show Source):
You can put this solution on YOUR website!find the inverse of: f(x)=3x+6
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1st: Interchange x and y to get:
x = 3y+6
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2nd: Solve for "y" to get the inverse.
3y = x-6
y = (1/3)x-2
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Cheers,
Stan H.
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Miscellaneous_Word_Problems/347783: Two cards are drawn from an ordinary deck of 52 cards. Find the probability of each event, showing your reasoning carefully.
a. Two aces.
b. Two red cards
c. Two red aces.
d. Two honor cards (A, K, Q, J, 10).
1 solutions
Answer 248654 by stanbon(57424) on 2010-09-24 21:21:54 (Show Source):
You can put this solution on YOUR website! Two cards are drawn from an ordinary deck of 52 cards. Find the probability of each event, showing your reasoning carefully.
a. Two aces.
P(2 aces) = 4C2/52C2
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b. Two red cards
P(2 red) = 26C2/52C2
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c. Two red aces.
d. Two honor cards (A, K, Q, J, 10).
P(2 honor) = 20C2/52C2
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Cheers,
Stan H.
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Miscellaneous_Word_Problems/347781: Jack usually mows his lawn in 5 hours, Marilyn can mow the same lawn in 4 hours. How much time would it take for them to mow the lawn together.
They could mow the lawn in _?_ hours if they worked together.
1 solutions
Answer 248652 by stanbon(57424) on 2010-09-24 21:18:13 (Show Source):
You can put this solution on YOUR website!Jack usually mows his lawn in 5 hours, Marilyn can mow the same lawn in 4 hours. How much time would it take for them to mow the lawn together.
They could mow the lawn in _?_ hours if they worked together.
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Jack rate: 1/5 job/hr
Marilyn rate: 1/4 job/hr
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Together rate: 1/x job/hr
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Equation:
rate + rate = together rate
1/5 + 1/4 = 1/x
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Multiply thru 20x to get:
4x + 5x = 20
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9x = 20
x = 20/9
x = 2 2/9 hrs (time to do the job together)
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Cheers,
Stan H.
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Probability-and-statistics/347745: Of a classroom filled with 20 students, 2 will be selected to stay after school and correct homework for extra credit. How many combinations are possible?
1 solutions
Answer 248631 by stanbon(57424) on 2010-09-24 19:25:06 (Show Source):
You can put this solution on YOUR website!Of a classroom filled with 20 students, 2 will be selected to stay after school and correct homework for extra credit. How many combinations are possible?
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20C2 = (20*19)/(1*2) = 190
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Cheers,
Stan H.
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