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Recent problems solved by 'stanbon'
stanbon answered: 57288 problems
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53010..53039 , 53040..53069 , 53070..53099 , 53100..53129 , 53130..53159 , 53160..53189 , 53190..53219 , 53220..53249 , 53250..53279 , 53280..53309 , 53310..53339 , 53340..53369 , 53370..53399 , 53400..53429 , 53430..53459 , 53460..53489 , 53490..53519 , 53520..53549 , 53550..53579 , 53580..53609 , 53610..53639 , 53640..53669 , 53670..53699 , 53700..53729 , 53730..53759 , 53760..53789 , 53790..53819 , 53820..53849 , 53850..53879 , 53880..53909 , 53910..53939 , 53940..53969 , 53970..53999 , 54000..54029 , 54030..54059 , 54060..54089 , 54090..54119 , 54120..54149 , 54150..54179 , 54180..54209 , 54210..54239 , 54240..54269 , 54270..54299 , 54300..54329 , 54330..54359 , 54360..54389 , 54390..54419 , 54420..54449 , 54450..54479 , 54480..54509 , 54510..54539 , 54540..54569 , 54570..54599 , 54600..54629 , 54630..54659 , 54660..54689 , 54690..54719 , 54720..54749 , 54750..54779 , 54780..54809 , 54810..54839 , 54840..54869 , 54870..54899 , 54900..54929 , 54930..54959 , 54960..54989 , 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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, >>Next
Probability-and-statistics/363019: The question was According to USA today, only 3.2% of all product coupons are redeemed. If a company distributes 15,000 coupons, what is the probability that:
a) more than 500 will be redeemed
b) fewer than 450 will be redeemed
1 solutions
Answer 258842 by stanbon(57361)   on 2010-10-28 10:59:47 (Show Source):
You can put this solution on YOUR website! The question was According to USA today, only 3.2% of all product coupons are redeemed. If a company distributes 15,000 coupons, what is the probability that:
a) more than 500 will be redeemed
z(500/15000) = z(0.0333) = (0.0333-0.032)/sqrt(0.032*0.968/15000) = 0.9046
P(p-hat > 0.3333) = P(z> 0.9046) = 0.1828
-------------------
b) fewer than 450 will be redeemed
z(450/15000) = (0.03-0.032)/0.001437 = -1.3918
---
P(p-hat < 450) = P(z < -1.3918) = 0.0820
=============================================
Cheers,
Stan H.
|
Probability-and-statistics/363020: An automatic filling machine fills bags of potato chips with 32 ounces of chips. The population deviation is 2 ounces. Convert weights of 31 ounces and 35 ounces to z-scores and sketch the graph of the weight distribution on a bell curve with weights of 31 and 35 indicated on the graph.
1 solutions
Answer 258794 by stanbon(57361) on 2010-10-28 09:03:01 (Show Source):
You can put this solution on YOUR website!An automatic filling machine fills bags of potato chips with 32 ounces of chips. The population deviation is 2 ounces. Convert weights of 31 ounces and 35 ounces to z-scores and sketch the graph of the weight distribution on a bell curve with weights of 31 and 35 indicated on the graph.
--------
z(31) = (31-32)/2 = -1/2
z(35) = (35-32)/2 = 3/2
--------------------
Can't show you the bell-shaped graph on this site.
Cheers,
Stan H.
|
Probability-and-statistics/363023: An English quiz contained two multiple choice questions that had three choices and five multiple choice questions with five choices. How many different ways is it possible to complete this test?
I have 28,125 and wanted to make sure this is correct. Got the answer by 3^2*5^5 = 28,125
1 solutions
Answer 258791 by stanbon(57361) on 2010-10-28 09:00:01 (Show Source):
You can put this solution on YOUR website!An English quiz contained two multiple choice questions that had three choices
3^2 = 9
---------------
and five multiple choice questions with five choices.
5^5 = 3125
--------
How many different ways is it possible to complete this test?
I have 28,125 and wanted to make sure this is correct. Got the answer by 3^2*5^5 = 28,125
====================
That looks good to me.
Cheers,
Stan H.
|
Probability-and-statistics/362835: It is estimated that 60% of all college graduates find a job in their chosen career field within one year of graduation. If six college graduates from the year 2009 are chosen at random, let x be the number who found a job within one year.
I have calculated the probabilities as follow: P(0) .004, P(1) .037, P(2) .138, P(3) .276, P(4) .311 P(5) .187 P(6) .047
a) what is the probability that fewer than 4 found a job in their field? I think it is .455
b) what is the probability that all of them found a job in their field? I think it is 1
c) Compute the mean, the expected number of job seekers (round to the appropriate whole number) I think it is 4 people
d) Compute the standard deviation of the number of successful job seekers. I think it is either 1.2 or 1.27
Any help with the doublechecks would be great. Thanks!
1 solutions
Answer 258665 by stanbon(57361) on 2010-10-27 21:45:51 (Show Source):
You can put this solution on YOUR website!It is estimated that 60% of all college graduates find a job in their chosen career field within one year of graduation. If six college graduates from the year 2009 are chosen at random, let x be the number who found a job within one year.
I have calculated the probabilities as follow: P(0) .004, P(1) .037, P(2) .138, P(3) .276, P(4) .311 P(5) .187 P(6) .047
----
Binomial Problem with n=6 and p = 0.6
a) what is the probability that fewer than 4 found a job in their field? I think it is .455
OK
------
b) what is the probability that all of them found a job in their field? I think it is 1
No.
P(x=6) = 0.06^6 = 0.04666 as you listed above.
------------------
c) Compute the mean, the expected number of job seekers (round to the appropriate whole number) I think it is 4 people
mean of binomial = np = 6*0.6 = 3.6
--------------------------
d) Compute the standard deviation of the number of successful job seekers. I think it is either 1.2 or 1.27
--
std of binomial is sqrt(npq) = sqrt(6*0.6*0.4)= 1.2
====================
Cheers,
Stan H.
|
Probability-and-statistics/362868: A manufacturer of desks has made 200 per week for the last year with a standard deviation (σ) of 16. Recently, a manufacturing change was made and now production is at 205 desks per week for the 50 randomly sampled days. Determine, at the 0.01 level of significance (α) using a 1 tail test if the change made any difference.
1 solutions
Answer 258664 by stanbon(57361) on 2010-10-27 21:32:10 (Show Source):
You can put this solution on YOUR website!A manufacturer of desks has made 200 per week for the last year with a standard deviation (σ) of 16.
Recently, a manufacturing change was made and now production is at 205 desks per week for the 50 randomly sampled days.
Determine, at the 0.01 level of significance (α) using a 1 tail test if the change made any difference.
----
Ho: u <= 200
Ha: u > 200
----
t(205) = (205-200)/[16/sqrt(50)] = 2.2097
---
p-value = P(t > 2.2097 when df = 49) = 0.0159
--------------------
Conclusion: Since the p-value is greater than
1%, fail to reject Ho.
--
The test does not show the change made a difference
at the 1% level of significance.
==============
Cheers,
Stan H.
========================================
|
Probability-and-statistics/362531: A digital camcorder repair service has set a goal not to exceed an average of 5 working days from
the time the unit is brought in to the time repairs are completed. A random sample of 12 repair
records showed the following repair times (in days): 9, 2, 5, 1, 5, 4, 7, 5, 11, 3, 7, 2. At α = .05 is
the goal being met? Repair
1 solutions
Answer 258597 by stanbon(57361) on 2010-10-27 19:44:43 (Show Source):
You can put this solution on YOUR website!A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair (in days) 9,2,5,1,5,4,7,5,11,3,7,2. At a=.05 is the goal being met?
----------------------------------------
Ho: u <= 5 (goal)
Ha: u > 5
---------------
x-bar = 5.08333333...
s = 2.998737....
t(5.08333) = (5.08333 - 5)/2.9987.../sqrt(12) = 0.096267
p-value for one-tail test with df=11 is P(0.096267 < t < 10) = 0.4625...
-----------------
Conclusion:
Since p-value is greater than 5% fail to reject Ho.
The goal is being met.
=========================
Cheers,
Stan H.
|
Probability-and-statistics/362591: Histograms for grouped data
Below are the times (in days) it takes for a sample of customers from Jack's computer store to pay their invoices.
19,15,43,39,35,31,27,22,18,26,26,21,21,21,17,17,13,13,41
Draw the histogram for these data using an initial class boundary of and a class width of . Note that you can add or remove classes from the figure. Label each class with its endpoints.
1 solutions
Answer 258578 by stanbon(57361) on 2010-10-27 19:35:26 (Show Source):
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Probability-and-statistics/362594: Albert and Elsie operate a very large and very busy U-pick strawberry patch. The mean number of quarts picked per day by the customers is 2100, with a standard deviation of 250. The distribution is normal. What is the number of quarts that separates the slowest 3% of the days from the other 97% of the days?
1 solutions
Answer 258574 by stanbon(57361) on 2010-10-27 19:34:17 (Show Source):
You can put this solution on YOUR website!Albert and Elsie operate a very large and very busy U-pick strawberry patch. The mean number of quarts picked per day by the customers is 2100, with a standard deviation of 250. The distribution is normal. What is the number of quarts that separates the slowest 3% of the days from the other 97% of the days?
-----
The z-value with a 3% left tail is -1.8808
----
Find x = zs+u
x = -1.8808*250+2100 = 1630 when rounded down to a whole number.
====================
Cheers,
Stan H.
|
Probability-and-statistics/362601: The following are the lengths of stay (in days) for a random sample of patients discharged from a particular hospital:
13,9,5,11,6,3,3,3,3,9,9,9,5,5
Draw the histogram for these data using an initial class boundary of , an ending class boundary of , and classes of equal width. Note that you can add or remove classes from the figure. Label each class with its endpoints.
1 solutions
Answer 258564 by stanbon(57361) on 2010-10-27 19:28:05 (Show Source):
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Probability-and-statistics/362674: Hello, Can someone please review my answer to the following question. Please help me.
A person spins the pointer and is awarded the amount indicated by the pointer. Circle with left half $2.00 the upper right quarter is $12 and the lower right quarter is $5.00 and it cost $6 to play the game. There are only 3 part to the circle
It costs $6 to play the game. Determine:
These are my answers I am not sure they are right
• The expectation of a person who plays the game.
E(x) = (1/2)2 + (1/4)12 + (1/4)5
E = 1+3+1.25 =5.25
•My Answer:The fair price to play the game. Wheel
Fair price = 6/3 = 2.00
1 solutions
Answer 258563 by stanbon(57361) on 2010-10-27 19:26:53 (Show Source):
You can put this solution on YOUR website!A person spins the pointer and is awarded the amount indicated by the pointer. Circle with left half $2.00 the upper right quarter is $12 and the lower right quarter is $5.00 and it cost $6 to play the game. There are only 3 part to the circle
It costs $6 to play the game. Determine:
These are my answers I am not sure they are right
• The expectation of a person who plays the game.
E(x) = (1/2)2 + (1/4)12 + (1/4)5
E = 1+3+1.25 =5.25
----
Your expected winnings are 5.25. But you
paid 6 to play the game. So your expected
"winnings" are really -0.75. You can expect
to lose 75 cents each time you play the game.
==============================================
To make the game "fair", the vendor should only
charge $5.25 to play the game.
=================================
Cheers,
Stan H.
|
Probability-and-statistics/362725: You have a coffee shop where you have averaged 50 cups of sales per hour with a standard deviation of 8 cups. Recently you did some advertising and now you want to see if the ads paid off. So you randomly sampled for 35 random hours and found that your sales went to 53 cups per hour. Use a one tail test and a 0.05 alpha to determine if the ads were worth the effort
1 solutions
Answer 258558 by stanbon(57361) on 2010-10-27 19:19:56 (Show Source):
You can put this solution on YOUR website!You have a coffee shop where you have averaged 50 cups of sales per hour with a standard deviation of 8 cups. Recently you did some advertising and now you want to see if the ads paid off. So you randomly sampled for 35 random hours and found that your sales went to 53 cups per hour. Use a one tail test and a 0.05 alpha to determine if the ads were worth the effort
----
Ho: u <= 50
Ha: u > 50 (means the ads worked)
-----
t(53) = (53-50)/[8/sqrt(35)] = 2.2185
-----------------
p-value = P(t> 2.2185 when df = 34) = 0.0167
------------
Conclusion:
Since the p-value is less than 5%, reject Ho.
The test supports the effectiveness of the ads.
===============================================
Cheers,
Stan H.
|
Probability-and-statistics/362698: I see where someone have asked the same question and it said that it had been answered but I could not find where it was answered. If it have been answered, can I be shown where; if not can it be answered for me?
1.) Assume that the population of heights of male college students is approximately normally distributed with the mean of 69 inches and standard deviation of 3.75 inches. Show all work.
(A). Find the proportion of male college students whose height is greater than 70 inches.
(B). Find the proportion of male college students whose heights is no more than 70 inches.
(2). Find the normal approximation for the binomial probability that x = 6, where n = 15 and p = 6.
(3). A set of data is normally distributed with a mean of 200 and standard deviation of 50.
*what would be the standard score for a score of 167?
*what percentage of scores is between 200 and 167?
*what would be the percentile rank for a score of 167?
1 solutions
Answer 258538 by stanbon(57361) on 2010-10-27 18:58:34 (Show Source):
You can put this solution on YOUR website!1.) Assume that the population of heights of male college students is approximately normally distributed with the mean of 69 inches and standard deviation of 3.75 inches. Show all work.
(A). Find the proportion of male college students whose height is greater than 70 inches.
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z(70) = (70-69)/3.75 = 0.2667
P(x > 70) = P(z > 0.2667) = 0.3949
(B). Find the proportion of male college students whose heights is no more than 70 inches.
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P(x<= 70) = 1-0.3949 = 0.6051
==============0.======================
(2). Find the normal approximation for the binomial probability that x = 6, where n = 15 and p = 6.
Find p(5.5 < x < 6.5) using mean = np=15*0.6 = 9
and std = sqrt(npq) = sqrt(9*0.4) = 1.8974
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z(5.5) = (5.5-9)/1.8974 = -1.8447
z(6.5) = (6.5-9)/1.8974 = -1.3176
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P(x=6) = P(-1.8447< z < -1.3176) = 0.0613
=================================================
(3). A set of data is normally distributed with a mean of 200 and standard deviation of 50.
*what would be the standard score for a score of 167?
z(167) = (167-200)/50 = -0.6600
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I'll leave the rest to you.
Cheers,
Stan H.
*what percentage of scores is between 200 and 167?
*what would be the percentile rank for a score of 167?
0 solutions
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Numbers_Word_Problems/362691: Solve,show work to help me understand
Mike said, "last week i ran 15 miles farther than bill." pete said "last week i ran one mile less than 3 times as far as bill" If mike and pete ran the same distance, how far did bill run?
1 solutions
Answer 258494 by stanbon(57361) on 2010-10-27 17:37:43 (Show Source):
You can put this solution on YOUR website!Mike said, "last week i ran 15 miles farther than bill." pete said "last week i ran one mile less than 3 times as far as bill" If mike and pete ran the same distance, how far did bill run?
------------------------------------
Equations:
m = b + 15
p = 3b -1
m = p
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Equation:
b + 15 = 3b -1
2b = 16
b = 8 miles (distance bill ran)
=================================
Cheers,
Stan H.
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Travel_Word_Problems/362666: how far can a man drive out into the country at the average rate of 60 mph and return over the same road at the average rate of 45 mph if he travels a total of 7 hours?
1 solutions
Answer 258491 by stanbon(57361) on 2010-10-27 17:32:45 (Show Source):
You can put this solution on YOUR website!How far can a man drive out into the country at the average rate of 60 mph and return over the same road at the average rate of 45 mph if he travels a total of 7 hours?
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Out bound DATA:
rate = 60 mph ; distance = x miles ; time = d/r = x/60 hrs.
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Return DATA:
rate = 45 mph ; distance = x miles ; time = d/r = x/45 hrs
------------------------
Equation:
time + time = 7
x/60 + x/45 = 7
----
45x + 60x = 7*60*45
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105x = 18900
x = 180 miles
================
Cheers,
Stan H.
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Finance/362660: An investment pays 3.4% compounded four times a year.
a. What is the annual growth factor?
Round to four decimal places.
b. What is the annual growth rate?
Round to two decimal places.
1 solutions
Answer 258470 by stanbon(57361) on 2010-10-27 16:45:02 (Show Source):
You can put this solution on YOUR website!An investment pays 3.4% compounded four times a year.
a. What is the annual growth factor?
Round to four decimal places.
(r/n)^4 = (1+(0.034/4))^4 = 1.0344
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b. What is the annual growth rate?
Round to two decimal places.
0.0344 = 3.44%
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Cheers,
Stan H.
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Numbers_Word_Problems/362574: The smaller of two numbers is 3 less than twice the larger number. If the larger number is decreased by 10, the result is the same as the smaller number. Find the numbers.
1 solutions
Answer 258422 by stanbon(57361) on 2010-10-27 14:21:14 (Show Source):
You can put this solution on YOUR website!The smaller of two numbers is 3 less than twice the larger number. If the larger number is decreased by 10, the result is the same as the smaller number. Find the numbers.
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larger: x
smaller: 2x-3
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Equation:
x-10 = 2x-3
x = -7 (larger)
2x-3 = -17 (smaller)
=========================
Cheers,
Stan H.
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Travel_Word_Problems/362534: Harry walked to the town at a speed of 4 miles/hr, and came back by bicycle at a speed of 12 miles/hr. The total time for to and fro journey was 5/2 hrs. Find the time taken by him to travel foot.
1 solutions
Answer 258386 by stanbon(57361) on 2010-10-27 12:58:15 (Show Source):
You can put this solution on YOUR website!Harry walked to the town at a speed of 4 miles/hr, and came back by bicycle at a speed of 12 miles/hr. The total time for to and fro journey was 5/2 hrs. Find the time taken by him to travel foot.
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Walk DATA:
rate = 4 mph ; distance = x miles ; time = x/4 hrs
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Bike DATA:
rate = 12 mph ; distance = x miles; time = 1/12 hrs
-----------------------
Equation:
time + time = 5/2
x/4 + x/12 = 5/2
---
3x + x = 30
4x = 30
x = 7.5 miles (
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Walk time = x/4 = (7.5/4)(60 min) = 112.5 min = 1 hr 52.5 min
==================
Cheers,
Stan H.
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Finance/362495: Betty has 10 more dimes than quarters. If she has $3.45, how many coins does she have?
1 solutions
Answer 258366 by stanbon(57361) on 2010-10-27 11:10:53 (Show Source):
You can put this solution on YOUR website!Betty has 10 more dimes than quarters. If she has $3.45, how many coins does she have?
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d = q + 10
10d + 25q = 345 cents
----
Substitute for "d" and solve for "q";
10(q+10) + 25q = 345
35q + 100 = 345
35q = 245
q = 7 (# of quarters)
---
Since d = q+10, d = 17 (# of dimes)
================
Cheers,
Stan H.
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Complex_Numbers/362494: i am trying to solve a problem with the format ax squared + bx + c and im not quit sure how to do it. can you help me?
1 solutions
Answer 258363 by stanbon(57361) on 2010-10-27 11:03:08 (Show Source):
You can put this solution on YOUR website! i am trying to solve a problem with the format ax squared + bx + c and im not quite sure how to do it. can you help me?
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Use the quadratic formula:
x = [-b +- sqrt(b^2-4ac)]/(2a)
to solve for "x".
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Cheers,
Stan H.
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Probability-and-statistics/362004: Hello, Could you please help me with this problem.
1. Essay; show all work. The property restoration company PuroServ is considering switching to new dehumidifiers. Their market research, considering the cost of the new machines and their efficiency, tells them that the switch would give them a 96% chance of making a $20,000 profit, a 1% chance of breaking even, and a 3% chance of losing $5,000. How much money does PuroServ expect to make with their new purchase?
Thank you.
1 solutions
Answer 258360 by stanbon(57361) on 2010-10-27 10:39:14 (Show Source):
You can put this solution on YOUR website!Their market research, considering the cost of the new machines and their efficiency, tells them that the switch would give them a 96% chance of making a $20,000 profit, a 1% chance of breaking even, and a 3% chance of losing $5,000. How much money does PuroServ expect to make with their new purchase?
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Let the random values be "gain" to the company
X-values are 20,000; 0; -5000
probabilities are: 0.96; 0.01; 0.03
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Expected gain = 20,000*0.96 + 0*0.01 - 5000*0.03 = $19050.00
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Cheers,
Stan H.
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Probability-and-statistics/362035: A TV special says there is a 10% incidence of sexually transmitted disease(STD)among all U.S. teens. A reporter fo the Rockville Bugle, looking for a story, surveyed 260 randomly chosen teenagers, and found that 39 had been to a clinic for STD treatment. Using a right tail test at a significance level of 0.05, the test statistic is:
A. 2.687
B. 2.758
C. 0.0256
D. 2.258
1 solutions
Answer 258359 by stanbon(57361) on 2010-10-27 10:33:48 (Show Source):
You can put this solution on YOUR website!A TV special says there is a 10% incidence of sexually transmitted disease(STD)among all U.S. teens. A reporter fo the Rockville Bugle, looking for a story, surveyed 260 randomly chosen teenagers, and found that 39 had been to a clinic for STD treatment. Using a right tail test at a significance level of 0.05, the test statistic is:
t(39/260) = (0.15-0.10)/[sqrt(0.10*0.90/260)] = 2.687
--------------
Cheeers,
Stan H.
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A. 2.687
B. 2.758
C. 0.0256
D. 2.258
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Probability-and-statistics/362292: am having problems understanding how to do the problem and I need to understand what steps needs to be done.
1.) Assume that the population of heights of male college students is approximately normally distributed with the mean of 69 inches and standard deviation of 3.75 inches. SHow all work.
(A). Find the proportion of male college students whose height is greater than 70 inches.
(B). Find the proportion of male college students whose heights is no more than 70 inches.
(2). Find the normal approximation for the binomial probability that x = 6, where n = 15 and p = 6.
(3). A set of data is normally distributed with a mean of 200 and standard deviation of 50.
*what would be the standard score for a score of 167?
*what percentage of scores is between 200 and 167?
*what would be the percentile rank for a score of 167?
1 solutions
Answer 258358 by stanbon(57361) on 2010-10-27 10:27:43 (Show Source):
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Probability-and-statistics/362424: When I throw 2 independent sets of three six sided dice (red A) and (blue B) at the same time what is the (p) that a score of either 9 10 or 11 will end up be the highest sum total from either set.
Mike mantan
1 solutions
Answer 258356 by stanbon(57361) on 2010-10-27 10:23:28 (Show Source):
You can put this solution on YOUR website!When I throw 2 independent sets of three six sided dice (red A) and (blue B) at the same time what is the (p) that a score of either 9 10 or 11 will end up be the highest sum total from either set.
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Sketch the sample space which has 2*3*6 = 36 ordered triples
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Count the number of triples whose sum is 9,10 or 11.
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Divide that number by 36 to get "p".
----------------
Cheers,
Stan H.
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Probability-and-statistics/362450: A fair die is tossed three times.What is the probability that, on at least two of thethrows, the face that appears on top is greater than 4?
1 solutions
Answer 258354 by stanbon(57361) on 2010-10-27 10:06:31 (Show Source):
You can put this solution on YOUR website!A fair die is tossed three times. What is the probability that, on at least two of the throws, the face that appears on top is greater than 4?
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Binomial Problem with n = 3, p = 2/6, and 2<=x<=3)
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Ans: 1 - binomcdf(3,1/3,1) = 0.2593
=================
Cheers,
Stan H.
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Probability-and-statistics/362457: 1).a set of 50 data values has a mean of 40 and a variance of 25.
a). Find the standard score (z) for a data value = 47.
b). Find the probability of a data value >47.
(2). Find the area under the standard normal curve:
I. to the right of z = 1.83
II. to the left of z = 1.83
1 solutions
Answer 258352 by stanbon(57361) on 2010-10-27 10:02:41 (Show Source):
You can put this solution on YOUR website!1).a set of 50 data values has a mean of 40 and a variance of 25.
a). Find the standard score (z) for a data value = 47.
z(47) = (47-40)/25 = 0.28
b). Find the probability of a data value > 47.
P(x> 47) = P(z > 0.28) = 0.3897
=====================================================
(2). Find the area under the standard normal curve:
I. to the right of z = 1.83
P(z > 1.83) = 0.0336
---
II. to the left of z = 1.83
P(z < 1.83) = 1-0.0336 = 0.9664
=================
Cheers,
Stan H.
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Polynomials-and-rational-expressions/362475: find the quotient and the remainder of expression
(3x^3+3x-2)/(x+1)
a. 3x^2+3x-4,6
b. 3x^2-2x,0
c. 3x^2+3x+4,2
d. 3x^2-3x+4,-6
1 solutions
Answer 258344 by stanbon(57361) on 2010-10-27 09:49:03 (Show Source):
You can put this solution on YOUR website!find the quotient and the remainder of expression
(3x^3+3x-2)/(x+1)
a. 3x^2+3x-4,6
b. 3x^2-2x,0
c. 3x^2+3x+4,2
d. 3x^2-3x+4,-6
-----------------------
Using synthetic division:
-1)....3...0...3...-4
........3...-3..6...|..-10
Quotient: 3x^2-3x+6
Remainder: -10
====================
Cheers,
Stan H.
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Systems-of-equations/362462: 6x^2-5x-1/12x^2-11x+2 is less than 0
1 solutions
Answer 258316 by stanbon(57361) on 2010-10-27 09:00:14 (Show Source):
You can put this solution on YOUR website!6x^2-5x-1/12x^2-11x+2 is less than 0
------
Factor:
[(6x+1)(x-1)]/[(3x-2)(4x-1)] < 0
------
Find the boundary values for the inequality
which are the solutions to the EQUALITY.
Those values are x = -1/6 and x = 1
Also take note of indeterminate values.
Those are x = 2/3 and x = 1/4
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Draw a number line and plot the boundary values:
....-1/6....1/4....2/3....1....
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To find the solutions for the inequality, test
a value from each of the three bounded intervals
in the inequality.
[(6x+1)(x-1)]/[(3x-2)(4x-1)] < 0
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If x = -1 you get: (-*-)/(-*-) < 0; this is false
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If x = 0 you get: (+*-)/(-*-) < 0: true, so solutions in (-1/6,1/4)
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Do the same for x = 1/3 , for x = 5/6, and x = 1
=====================================================
Cheers,
Stan H.
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