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Recent problems solved by 'stanbon'
stanbon answered: 57894 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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54990..55019 , 55020..55049 , 55050..55079 , 55080..55109 , 55110..55139 , 55140..55169 , 55170..55199 , 55200..55229 , 55230..55259 , 55260..55289 , 55290..55319 , 55320..55349 , 55350..55379 , 55380..55409 , 55410..55439 , 55440..55469 , 55470..55499 , 55500..55529 , 55530..55559 , 55560..55589 , 55590..55619 , 55620..55649 , 55650..55679 , 55680..55709 , 55710..55739 , 55740..55769 , 55770..55799 , 55800..55829 , 55830..55859 , 55860..55889 , 55890..55919 , 55920..55949 , 55950..55979 , 55980..56009 , 56010..56039 , 56040..56069 , 56070..56099 , 56100..56129 , 56130..56159 , 56160..56189 , 56190..56219 , 56220..56249 , 56250..56279 , 56280..56309 , 56310..56339 , 56340..56369 , 56370..56399 , 56400..56429 , 56430..56459 , 56460..56489 , 56490..56519 , 56520..56549 , 56550..56579 , 56580..56609 , 56610..56639 , 56640..56669 , 56670..56699 , 56700..56729 , 56730..56759 , 56760..56789 , 56790..56819 , 56820..56849 , 56850..56879 , 56880..56909 , 56910..56939 , 56940..56969 , 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 , 57360..57389 , 57390..57419 , 57420..57449 , 57450..57479 , 57480..57509 , 57510..57539 , 57540..57569 , 57570..57599 , 57600..57629 , 57630..57659 , 57660..57689 , 57690..57719 , 57720..57749 , 57750..57779 , 57780..57809 , 57810..57839 , 57840..57869 , 57870..57899, >>NextTravel_Word_Problems/524073: Hello, can you please solve and explain this question while showing me every step. thank you
A boy walks 25 miles East then 8 miles North,how far is he from the starting point?
1 solutions
Answer 347535 by stanbon(57967)   on 2011-10-31 18:56:38 (Show Source):
You can put this solution on YOUR website!A boy walks 25 miles East then 8 miles North,how far is he from the starting point?
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Sketch the problem on an x/y coordinate system.
---
Start the boy's journey at (0,0)
He moves 25 East to (25,0)
He then moves 8 North to (25,8)
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Find the distance from (0,0) to (25,8)
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d = sqrt[25^2+8^2] = sqrt[625+64] = sqrt[689] = 26.25 miles
========================================================
Cheers,
Stan H.
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logarithm/524071: Use the properties of logarithms to condense each logarithmic expression. Write the expression as a single logarithm whose coefficient is 1. the problem: log base 8 (x+5)+1/3 log base 8x
1 solutions
Answer 347533 by stanbon(57967) on 2011-10-31 18:50:56 (Show Source):
You can put this solution on YOUR website!Write the expression as a single logarithm whose coefficient is 1. the problem: log base 8 (x+5)+1/3 log base 8x
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= log8[(x+5)*x^(1/3)]
OR---
= log8[x^(4/3)+5x^(1/3)]
=========================
Cheers,
Stan H.
|
Rational-functions/523899: Hello. Please help with this.
I have to write the equation of y+ x^2 + 6x + 5 in vertex form. Showing all work.
I appreciate your help!
1 solutions
Answer 347466 by stanbon(57967) on 2011-10-31 13:32:30 (Show Source):
You can put this solution on YOUR website!y = x^2 + 6x + 5 in vertex form
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y = (x^2+6x+?) + 5-?
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Complete the square:
y = (x^2+6x+9) + 4 - 9
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y = (x+3)^2-5
---------------------
Cheers,
Stan H.
|
Rational-functions/523894: Could you please help. I need to write the equation of the parabola in vertex form.
1. The parabola has it's vertex at the origin. Write the equation of the parabola if it also passes through the pooint (-1,2)
Thank you so much!
1 solutions
Answer 347465 by stanbon(57967) on 2011-10-31 13:29:17 (Show Source):
You can put this solution on YOUR website!Could you please help. I need to write the equation of the parabola in vertex form.
1. The parabola has it's vertex at the origin. Write the equation of the parabola if it also passes through the point (-1,2)
----
Plot those 2 points and you will see the parabola
must be opening toward the top.
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Vertex Form: y = a(x-h)+k where (h,k) is the vertex.
----
Using your two given points solve for "a":
2 = a(-1-0)^2+0
---
2 = a*1
a = 2
--------
Equation:
y = 2(x-0)^2 + 0
y = 2x^2
=========================
Cheers,
Stan H.
===============================
|
Probability-and-statistics/523804: The SAT Quantitative Test has a population mean of 500 and a population standard deviation of 100. Suppose you randomly select a sample of 25 participants and administer the SAT Quantitative Test to each participant.
a. What would you expect the mean of this sample to be? _________________
b. What is the standard error of the mean for samples of size 25 from this population?
____________________
c. What is the probability that this sample will be one of the samples that will have a mean SAT Quantitative score greater than 530?
____________________
d. What is the probability that this sample will be one of the samples that will have a mean SAT Quantitative score less than 480?
____________________
e. What is the probability that this sample will be one of the samples that will have a mean SAT Quantitative score between 485 and 525?
_____________________
f. Suppose we obtain a mean SAT Quantitative Score of 515. What is the 95% confidence interval for these data?
_____________________
g. Supposed we obtain a mean SAT Quantitative Score of 487. What is the 80% confidence interval for these data?
1 solutions
Answer 347464 by stanbon(57967) on 2011-10-31 12:57:40 (Show Source):
You can put this solution on YOUR website!The SAT Quantitative Test has a population mean of 500 and a population standard deviation of 100.
Suppose you randomly select a sample of 25 participants and administer the SAT Quantitative Test to each participant.
----------------------------
a. What would you expect the mean of this sample to be?:::500
--------------
b. What is the standard error of the mean for samples of size 25 from this population?:::100/sqrt(25) = 20
--------------
c. What is the probability that this sample will be one of the samples that will have a mean SAT Quantitative score greater than 530?
--
t(530) = (530-500)/20 = 3/2
P(x-bar) = P(t > 3/2 when df = 24) = 0.0733
--------------
d. What is the probability that this sample will be one of the samples that will have a mean SAT Quantitative score less than 480?
---
t(480) = (480-500)/20 = -1
P(x-bar > 480) = P(t < -1 when df = 24) = 0.1636
--------------
e. What is the probability that this sample will be one of the samples that will have a mean SAT Quantitative score between 485 and 525?
Find the t-scores and find that probability between them_____________________
f. Suppose we obtain a mean SAT Quantitative Score of 515. What is the 95% confidence interval for these data?
x-bar = 515
ME = 2.06*20 = 41.28
95% CI: 515-41.28 < u < 515+41.28
_____________________
g. Supposed we obtain a mean SAT Quantitative Score of 487. What is the 80% confidence interval for these data?
487-ME < u < 487+ME
ME = 1.32*20 = 26.36
=========================
Cheers,
Stan H.
============
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Probability-and-statistics/523870: If i have a 1 in 35.1 chance of winning off one powerball ticket. What is the probability that i would win something if i bought 10 tickets?
1 solutions
Answer 347463 by stanbon(57967) on 2011-10-31 12:38:44 (Show Source):
You can put this solution on YOUR website!If i have a 1 in 35.1 chance of winning off one powerball ticket. What is the probability that i would win something if i bought 10 tickets?
------
P(win) = 1/36.1 = 10/361
----
P(lose) = 9/361
----
P(win at least one with 10 ticks) = 1 - P(lose with all 10)
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= 1-(9/361)^10
---
(9/361)^10 = 2.39x10^-25
---
P(win at least one with 10 ticks) = 1- (2.39x10^-25)
=======================================================
Cheers,
Stan H.
===========
|
Probability-and-statistics/523868: For each question on a multiple choice test, there are 5 possible answers of which exactly one is correct.
If a student selects answers at random, give the probability that the first question answered correctly is question 4.
1 solutions
Answer 347461 by stanbon(57967) on 2011-10-31 12:32:12 (Show Source):
You can put this solution on YOUR website!For each question on a multiple choice test, there are 5 possible answers of which exactly one is correct.
If a student selects answers at random, give the probability that the first question answered correctly is question 4.
---
Let "w" be wrong; Let "c" be correct.
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P(wwwc) = (4/5)^3*(1/5) = 0.1024
------
Cheers,
Stan H.
|
Probability-and-statistics/523819: hello
I have some question and I need the answers today because I have an assignment
Q1: This homework question:
When a pizza restaurant's delivery process is operating effectively
When a pizza restaurant's delivery process is operating effectively, pizzas are delivered in an average of 45 minutes with a standard deviation of 6 minutes. To monitor its delivery process, the restaurant randomly selects five pizzas each night and records their delivery times.
a. For the sake of argument, assume that the population of all delivery times on a given evening is normally distributed with a mean of = 45 minutes and a standard deviation of = 6 minutes. (That is we assume that the delivery process is operating effectively)
a. Find the mean of the population of all possible sample means and the standard deviation of the population of all possible sample means and Calculate an interval containing 99.73% of all possible sample means.
b. Suppose that the mean of the five sampled delivery times on a particular evening is x� = 55 minutes. Using the interval that was calculated in a(4), what would you conclude about whether the restaurant's delivery process is operating effectively? Why?
Q2: Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. A key measure of customer satisfaction is the response (on a scale from 1 to 10) to the question: “Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?” Here, a response of 9 or 10 represents “customer delight.”
Historically, the percentage of Bank of America customers expressing customer delight has been 48%. Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight. The survey finds that 189 of 350 randomly selected Bank of America customers express customer delight. If, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing a sample proportion greater than or equal to 189/350 = .54.
1 solutions
Answer 347442 by stanbon(57967) on 2011-10-31 10:31:55 (Show Source):
You can put this solution on YOUR website!a. For the sake of argument, assume that the population of all delivery times on a given evening is normally distributed with a mean of = 45 minutes and a standard deviation of = 6 minutes. (That is we assume that the delivery process is operating effectively)
a. Find the mean of the population of all possible sample means and the standard deviation of the population of all possible sample means and Calculate an interval containing 99.73% of all possible sample means.
---
mean of the sample means = 45
std of the sample means = 6/sqrt(5) = 2.6833
---
1-0.9973 = 0.0027
(1/2)0.0027 = 0.0014
---
Find the t value with a left tail of 0.0014 when df = 4
t = invT(0.0014,4) = -6.56
Find the correspondiing raw score value using x = ts+u
x = -6.56*2.6833+45 = 27.41
OR x = +6.56*6833+45 = 62.60
Interval with 99.73% of sample means = (27.41,62.60)
=================================================================
b. Suppose that the mean of the five sampled delivery times on a particular evening is x� = 55 minutes. Using the interval that was calculated in a(4), what would you conclude about whether the restaurant's delivery process is operating effectively? Why?
---
55 is in the 99.73% confidence window
===================================================
Q2: Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. A key measure of customer satisfaction is the response (on a scale from 1 to 10) to the question: “Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?” Here, a response of 9 or 10 represents “customer delight.”
Historically, the percentage of Bank of America customers expressing customer delight has been 48%. Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight. The survey finds that 189 of 350 randomly selected Bank of America customers express customer delight. If, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing a sample proportion greater than or equal to 189/350 = .54.
---
z(0.54) = (0.54-0.48)/sqrt[0.48*0.52/350] = 2.2468
----
P(p-hat > 0.54) = P(z > 2.2468) = normalcdf(2.2468,100) = 0.0123
========================
Cheers,
Stan H.
========================
|
Probability-and-statistics/523757: Give probabilities for the following
age :.....<20....20-40.....>40(age group)
male:.....168....340.......170
female:...208....290.......160
1.Using the relative frequency of occurrence approach, what is the probability that a customer is a male?
2.What is the probability that a customer is 20 – 40 years old?
3.What is the joint probability of a customer being 20 – 40 years old and a male?
4.Calculate the probability that a customer selected from the male customers would be less than 20 years old. Calculate the same for the female customers. Does it appear that gender is independent of age? Support your answer with calculations and reasons.
1 solutions
Answer 347436 by stanbon(57967) on 2011-10-31 10:07:08 (Show Source):
You can put this solution on YOUR website!Give probabilities for the following:
age :.....<20....20-40.....>40(age group)..Totals
male:.....168....340.......170.............678
female:...208....290.......160.............658
Totals:...376....630.......330............1336
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1.Using the relative frequency of occurrence approach, what is the probability that a customer is a male?::::678/1336
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2.What is the probability that a customer is 20 – 40 years old?630/1336
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3.What is the joint probability of a customer being 20 – 40 years old and a male?:::340/1336
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4.Calculate the probability that a customer selected from the male customers would be less than 20 years old.:::168/376
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Calculate the same for the female customers.:::208/376
Does it appear that gender is independent of age?::Yes
Support your answer with calculations and reasons.:::Expected probability
for both would be (188/376). The observed probs are quite different.
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Cheers,
Stan H.
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Geometry_Word_Problems/523790: Find the obtuse angle between the y- axis and the line with a slope of m= -1/4?
1 solutions
Answer 347432 by stanbon(57967) on 2011-10-31 09:24:03 (Show Source):
You can put this solution on YOUR website!Find the obtuse angle between the y- axis and the line with a slope of m= -1/4?
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Plot the point (4,-1) and (0,0)
Draw a line thru them so you can see what is happening.
---
invtan(-1/4) = -14.04 degrees is the acute angle.
----
180-14.04 = 165.96 degrees is the obtuse angle.
================
Cheers,
Stan H.
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Geometry_Word_Problems/523796: Find the equation of the line through (2,3) and parallel to 3x-2y=7?
1 solutions
Answer 347429 by stanbon(57967) on 2011-10-31 09:16:27 (Show Source):
You can put this solution on YOUR website!Find the equation of the line through (2,3) and parallel to 3x-2y=7?
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Find the slope of the given equation:
3x-2y = 7
2y = 3x-7
y = (3/2)x - (7/2)
----
slope = 3/2
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Any line parallel to it must have slope = 3/2
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Form: y = mx + b
m = 3/2 and y = 3 when x = 2
Solve for "b":
3 = (3/2)*2 + b
b = 0
----
Equation:
y = (3/2)x
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cheers,
Stan H.
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Travel_Word_Problems/523791: It took cindy 2 hours to bike from abbott to benson at a constant speed. the return trip took only 1.5 hours because she increased her speed by 6 km/h. how far apart are abbott and benson?
I would like to know how you would solve a problem like this using the distance formula: speed*time=distance
1 solutions
Answer 347425 by stanbon(57967) on 2011-10-31 09:07:52 (Show Source):
You can put this solution on YOUR website!It took cindy 2 hours to bike from abbott to benson at a constant speed. the return trip took only 1.5 hours because she increased her speed by 6 km/h. how far apart are abbott and benson?
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A to B DATA:
time = 2 hrs ; rate = r kmh ; distance = r*t = 2r km
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B to A DATA:
time = 1.5 hrs; r+6 kmh ; distance = r*t = 1.5(r+6) km
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Equation:
distance = distance
2r = 1.5(r+6)
2r = 1.5r + 9
0.5r = 9
r = 18 kmh
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Distance from A to B = 2r = 2*18 = 36 km
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=======================================
cheers,
Stan H.
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Age_Word_Problems/523786: The Eastside Nursery ordered 27 trees. Some of the trees were elms, costing $17 each. The remainder was maples at $11 each. The total cost of the trees was $375. Find the number of elms and the number of maples.
1 solutions
Answer 347421 by stanbon(57967) on 2011-10-31 08:57:32 (Show Source):
You can put this solution on YOUR website!The Eastside Nursery ordered 27 trees. Some of the trees were elms, costing $17 each. The remainder was maples at $11 each. The total cost of the trees was $375. Find the number of elms and the number of maples.
---------------------------------
Equations:
e + m = 27
17e+11m = 375
---
Multiply thru the 1st equation by 17:
17e + 17m = 17*27
17e + 11m = 375
----
Subtract and solve for "m":
6m = 84
m = 14 (# of maple trees purchased)
----
Solve for "e":
e + m = 27
e + 14 = 27
e = 13 (# of elm trees purchased)
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Cheers,
Stan H.
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Problems-with-consecutive-odd-even-integers/523779: Find two consecutive odd integers such that their sum is 52 and their product is 675.
Numbers have to be entered in increasing order. Thank for the help!
1 solutions
Answer 347417 by stanbon(57967) on 2011-10-31 08:48:49 (Show Source):
You can put this solution on YOUR website!Find two consecutive odd integers such that their sum is 52 and their product is 675.
Numbers have to be entered in increasing order.
----------
1st: 2x-1
2nd: 2x+1
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Equation:
Sum: 4x = 52
Prod: (4x^2-1) = 675
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x = 52/4 = 13
----
Check:
4x^2-1 = 4*13^2-1 = 4*169-1 = 676-1 = 675
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Ans:
1st = 2x-1 = 2*13-1 = 25
2nd = 2x+1 = 2*13+1 = 27
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Cheers,
Stan H.
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Probability-and-statistics/523759: The amounts of money requested in home loan applications at the federal savings are approximately normally distributed with a mean of $70000 and a standard deviation of $20000. a loan application is received. what is the probability that:
1. the amount requested is $80 000 or more
2. twenty percent of the loans are larger than what amount
1 have tried to work the first question
80000 - 70000 / 20000 = 0,05
= 0.5 - 0,05
= 1.65 am stuck
1 solutions
Answer 347416 by stanbon(57967) on 2011-10-31 08:20:22 (Show Source):
You can put this solution on YOUR website!The amounts of money requested in home loan applications at the federal savings are approximately normally distributed with a mean of $70000 and a standard deviation of $20000. a loan application is received. what is the probability that:
1. the amount requested is $80 000 or more
1 have tried to work the first question
z(80000) = (80000 - 70000) / 20000 = 0.5
----
P(x >= 80000) = P(z>= 0.5) = 0.3085
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2. twenty percent of the loans are larger than what amount
Find the z-value with a left tail of 80%
invNorm(0.8) = 0.8416
---
Find the corresponding loan value using x = zs+u
x = 0.8416*20000+70000 = $86832.42
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Cheers,
Stan H.
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Money_Word_Problems/523675: How long it will take for an investment of 2000 dollars to double in value if the interest rate is 10 percent per year, compounded continuously?
1 solutions
Answer 347367 by stanbon(57967) on 2011-10-30 21:52:51 (Show Source):
You can put this solution on YOUR website!How long it will take for an investment of 2000 dollars to double in value if the interest rate is 10 percent per year, compounded continuously?
---
A(t) = P*e^(rt)
----
4000 = 2000*e^(0.1*t)
2 = e^ (0.1t)
---
Take the natural log of both sides to get:
0.1t = ln(2)
---
t = 0.6931/0.1
---
t = 6.931 years
=======================
Cheers,
Stan H.
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Rational-functions/523656: My homework is on story problems and i just dont know how to set it up niether do my parents.
Question:
YOur aunt and uncle have been visiting at your home. Five minutes after they drive away, you realize that they forgot thier bags. You happen to know that they drive slowly, so you get in your car and drive to catch up to them. Your average speed is 10 mph faster than thier average speed,and you catch up with them in 25 minutes. How fast did you drive?
Verbal Model- Your speed = Their Speed + 10 mph
Your speed x Your time in hours = Their speed x Their speed in hours
1 solutions
Answer 347364 by stanbon(57967) on 2011-10-30 21:39:39 (Show Source):
You can put this solution on YOUR website!YOur aunt and uncle have been visiting at your home.
Five minutes after they drive away, you realize that they forgot thier bags.
----
You happen to know that they drive slowly, so you get in your car and drive to catch up to them.
-----
Your average speed is 10 mph faster than their average speed,and you catch up with them in 25 minutes.
--------
How fast did you drive?
----
"Your" DATA:
rate = r+10 mph ; time = 25 min = 5/12 hr ; distance = r*t = (5/12)(r+10) miles
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"Their" DATA:
rate = r mph ; time = 30 min = 1/2 hr ; distance = r*t = (1/2)r miles
--------
Equation:
your distance = their distance
(5/12)(r+10) = (1/2)r
---
Multiply thru by 12 to get:
5(r+10) = 6r
5r+50 = 6r
r = 50
r = 10 mph (Their driving rate)
r+10 = 20 mph (Your driving rate)
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Cheers,
Stan H.
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Money_Word_Problems/523646: Alex, Sandra, and Mary were counting their pennies. Alex and Sandra together had 31. Sandra and Mary together had 19. Alex and Mary together had 24. How may pennies did each student have?
1 solutions
Answer 347361 by stanbon(57967) on 2011-10-30 21:20:30 (Show Source):
You can put this solution on YOUR website!Alex, Sandra, and Mary were counting their pennies. Alex and Sandra together had 31. Sandra and Mary together had 19. Alex and Mary together had 24. How may pennies did each student have?
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Equations:
a + 0 + s = 31
0 + m + s = 19
a + m + 0 = 24
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Use any method you know to get:
a = 18
m = 6
s = 13
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Cheers,
Stan H.
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Linear_Equations_And_Systems_Word_Problems/523649: sue bought 4 packs of cookies and 3 sacks of potato chips for 5.01. Later she bought 3 pack of cookies and 5 sacks of potato chips for 5.27. what is the price of a pack of cookies?
1 solutions
Answer 347357 by stanbon(57967) on 2011-10-30 21:10:56 (Show Source):
You can put this solution on YOUR website!sue bought 4 packs of cookies and 3 sacks of potato chips for 5.01. Later she bought 3 pack of cookies and 5 sacks of potato chips for 5.27. what is the price of a pack of cookies?
------------
Equations:
4c + 3p = 5.01
3c + 5p = 5.27
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Multiply thru 1st equation by 3
Multiply thru 2nd equation by 4
----------------------
12c + 9p = 3*5.01
12c + 20p = 4*5.27
----------------------------
Subtract and solve for "p":
11p = 6.05
---
p = $0.55 (cost of potato chips)
----
Solve for "c":
3c + 5p = 5.27
3c + 5*0.55 = 5.27
3c = 2.52
c = $0.84 (cost of cookies)
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Cheers,
Stan H.
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Numbers_Word_Problems/523638: the sum of three numbers is 237. the second number is 9 less than two times the first number. the third number is 6 more than three times the first number. what are the three numbers?
1 solutions
Answer 347353 by stanbon(57967) on 2011-10-30 20:54:07 (Show Source):
You can put this solution on YOUR website!the sum of three numbers is 237. the second number is 9 less than two times the first number. the third number is 6 more than three times the first number. what are the three numbers?
------
Equations:
x + y + z = 237
y = 2x-9
z = 3x+6
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Substitute for "y" and for "x"; then solve for "x":
x + (2x-9) + (3x+6) = 237
---
6x - 3 = 237
6x = 240
x = 40
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y = 2x-9 = 80-9 = 71
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z = 3x+6 = 126
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Cheers,
Stan H.
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Linear_Algebra/523635: " what is the slope of  "
"i have never understood this so please explain thank you."
1 solutions
Answer 347349 by stanbon(57967) on 2011-10-30 20:46:04 (Show Source):
You can put this solution on YOUR website!" what is the slope of "
"i have never understood this so please explain thank you."
----
Keep this in mind: The slope is the change in "y" when the
change in "x" is 1.
----
Try it:
y = 2x+3
---
Let x = 0, then y = 3
Let x = 1, then y = 2*1+3 = 5
----
y increased by 2 when x increased by 1
The slope is 2
----
Try it with y = -3x + 5
and you will see that the slope is -3.
Good Luck
===========================================
Cheers,
Stan H.
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Numbers_Word_Problems/523623: Please help me solve this problem.If 6 men can do a piece of work in 14 days, how many men are needed to do the work in 21 days?
1 solutions
Answer 347347 by stanbon(57967) on 2011-10-30 20:42:07 (Show Source):
You can put this solution on YOUR website!Please help me solve this problem.If 6 men can do a piece of work in 14 days, how many men are needed to do the work in 21 days?
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# of days and # of men are inversely related.
---
m = k/d
----
Solve for "k" using "6 men can do a piece of work in 14 days".
6 = k/14
k = 6*14 = 84
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Equation:
m = 84/d
---
Solve: How many men are needed to do the work in 21 days?
m = 84/21
m = 4
# of men needed is 4
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Cheers,
Stan H.
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logarithm/523624: use log 2=.301, log 3= .699 to compute log (1/20) to the nearest three decimal places.
1 solutions
Answer 347344 by stanbon(57967) on 2011-10-30 20:36:59 (Show Source):
You can put this solution on YOUR website!use log 2=.301, log 5= .699 to compute log (1/20) to the nearest three decimal places.
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1/20 = 1/(2^2*5)
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log(1/20) = log(20)^-1 = -log(2^2*5)
----
= -[2log(2) + log(5)]
----
= -[2*0.301 + 0.699]
----
= -1.3010
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Cheers,
Stan H.
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Probability-and-statistics/523606: Find the lower limit on a 95% confidence interval for the height of men in the U. S. (Consider each man a sample of size 1.)
Mean = 70.21 inches
Std. Dev. = 1.02 inches
1 solutions
Answer 347337 by stanbon(57967) on 2011-10-30 20:16:19 (Show Source):
You can put this solution on YOUR website! Find the lower limit on a 95% confidence interval for the height of men in the U. S. (Consider each man a sample of size 1.)
Sample Mean = 70.21 inches
Sample Std. Dev. = 1.02 inches
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ME = 1.96*1.02/sqrt(1) = 1.9992
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95% CI: 70.21-1.9992 < u < 70.21+1.9992
----
Lower Limit: 70.21-1.9992 = 68.21
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Cheers,
Stan H.
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