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Recent problems solved by 'stanbon'
stanbon answered: 57239 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 , 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45090..45119 , 45120..45149 , 45150..45179 , 45180..45209 , 45210..45239 , 45240..45269 , 45270..45299 , 45300..45329 , 45330..45359 , 45360..45389 , 45390..45419 , 45420..45449 , 45450..45479 , 45480..45509 , 45510..45539 , 45540..45569 , 45570..45599 , 45600..45629 , 45630..45659 , 45660..45689 , 45690..45719 , 45720..45749 , 45750..45779 , 45780..45809 , 45810..45839 , 45840..45869 , 45870..45899 , 45900..45929 , 45930..45959 , 45960..45989 , 45990..46019 , 46020..46049 , 46050..46079 , 46080..46109 , 46110..46139 , 46140..46169 , 46170..46199 , 46200..46229 , 46230..46259 , 46260..46289 , 46290..46319 , 46320..46349 , 46350..46379 , 46380..46409 , 46410..46439 , 46440..46469 , 46470..46499 , 46500..46529 , 46530..46559 , 46560..46589 , 46590..46619 , 46620..46649 , 46650..46679 , 46680..46709 , 46710..46739 , 46740..46769 , 46770..46799 , 46800..46829 , 46830..46859 , 46860..46889 , 46890..46919 , 46920..46949 , 46950..46979 , 46980..47009 , 47010..47039 , 47040..47069 , 47070..47099 , 47100..47129 , 47130..47159 , 47160..47189 , 47190..47219 , 47220..47249 , 47250..47279 , 47280..47309 , 47310..47339 , 47340..47369 , 47370..47399 , 47400..47429 , 47430..47459 , 47460..47489 , 47490..47519 , 47520..47549 , 47550..47579 , 47580..47609 , 47610..47639 , 47640..47669 , 47670..47699 , 47700..47729 , 47730..47759 , 47760..47789 , 47790..47819 , 47820..47849 , 47850..47879 , 47880..47909 , 47910..47939 , 47940..47969 , 47970..47999 , 48000..48029 , 48030..48059 , 48060..48089 , 48090..48119 , 48120..48149 , 48150..48179 , 48180..48209 , 48210..48239 , 48240..48269 , 48270..48299 , 48300..48329 , 48330..48359 , 48360..48389 , 48390..48419 , 48420..48449 , 48450..48479 , 48480..48509 , 48510..48539 , 48540..48569 , 48570..48599 , 48600..48629 , 48630..48659 , 48660..48689 , 48690..48719 , 48720..48749 , 48750..48779 , 48780..48809 , 48810..48839 , 48840..48869 , 48870..48899 , 48900..48929 , 48930..48959 , 48960..48989 , 48990..49019 , 49020..49049 , 49050..49079 , 49080..49109 , 49110..49139 , 49140..49169 , 49170..49199 , 49200..49229 , 49230..49259 , 49260..49289 , 49290..49319 , 49320..49349 , 49350..49379 , 49380..49409 , 49410..49439 , 49440..49469 , 49470..49499 , 49500..49529 , 49530..49559 , 49560..49589 , 49590..49619 , 49620..49649 , 49650..49679 , 49680..49709 , 49710..49739 , 49740..49769 , 49770..49799 , 49800..49829 , 49830..49859 , 49860..49889 , 49890..49919 , 49920..49949 , 49950..49979 , 49980..50009 , 50010..50039 , 50040..50069 , 50070..50099 , 50100..50129 , 50130..50159 , 50160..50189 , 50190..50219 , 50220..50249 , 50250..50279 , 50280..50309 , 50310..50339 , 50340..50369 , 50370..50399 , 50400..50429 , 50430..50459 , 50460..50489 , 50490..50519 , 50520..50549 , 50550..50579 , 50580..50609 , 50610..50639 , 50640..50669 , 50670..50699 , 50700..50729 , 50730..50759 , 50760..50789 , 50790..50819 , 50820..50849 , 50850..50879 , 50880..50909 , 50910..50939 , 50940..50969 , 50970..50999 , 51000..51029 , 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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 , 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, >>NextProbability-and-statistics/424083: Suppose X is a normally distributed random variable with the mean=10.1 and standard deviation=3.6. Find each of the following probabilities:
P(10.1<=x<=15.6)
and
P(x>=8)
1 solutions
Answer 295733 by stanbon(57307) on 2011-03-19 09:11:17 (Show Source):
You can put this solution on YOUR website!Suppose X is a normally distributed random variable with the mean=10.1 and standard deviation=3.6.
Find each of the following probabilities:
P(10.1<=x<=15.6)
---
z(10.1) = (10.1-10.1)/3.6 = 0
z(15.6) = (15.6-10.1)/3.6 = 5.5/3.6 = 1.5278
---
P(10.1 <= x <= 15.6) = P(0<= z <=1.5278) = 0.4367
===============================
and
P(x>=8)
z(8) = (8-10.1)/3.6 = -0.58333
P(x >= 8) = P(z >= -0.58333) = 0.7202
================
Cheers,
Stan H.
================
|
Probability-and-statistics/424087: suppose 5 cards are drawn at random from a pack of 52 cards .If all cards are red what is the probability that all of them are hearts? 1 solutions
Answer 295730 by stanbon(57307) on 2011-03-19 09:02:02 (Show Source):
You can put this solution on YOUR website!suppose 5 cards are drawn at random from a pack of 52 cards .
If all cards are red what is the probability that all of them are hearts?
---------------------
P(a card is heart) = 1/2
P(a card is not heart) = 1/2
-----
P(x = 5 hearts) = (1/2)^5 = 1/32
==================
Cheers,
Stan H.
|
Quadratic_Equations/424086: Translate the following into a quadratic equation, and solve it:
The length of a rectangular garden is two more than its width; if the area of the garden is 35 square meters, what are its dimensions?
1 solutions
Answer 295727 by stanbon(57307) on 2011-03-19 08:56:39 (Show Source):
You can put this solution on YOUR website!Translate the following into a quadratic equation, and solve it:
The length of a rectangular garden is two more than its width; if the area of the garden is 35 square meters, what are its dimensions?
------
Let width be "x"; then length is "x+2".
Area = (length)(width)
35 = (x+2)(x)
x^2+2x-35 = 0
Factor:
(x+7)(x-5) = 0
Positive solution:
x = 5 meters (width)
x+2 = 7 meters (length)
===========================
Cheers,
Stan H.
---------------------
|
Linear_Equations_And_Systems_Word_Problems/424085: Height and Weight Suppose that the weight of a
person is directly proportional to the cube of the
person’s height. If one person weighs twice as much
as a (similarly proportioned) second person, by what
factor is the heavier person’s height greater than the
shorter person’s height?
1 solutions
Answer 295725 by stanbon(57307) on 2011-03-19 08:53:15 (Show Source):
You can put this solution on YOUR website!Height and Weight
---------------------------------
Suppose that the weight of a
person is directly proportional to the cube of the
person’s height.
--
W = k*H^3
----------------------------------------
If one person weighs twice as much
as a (similarly proportioned) second person, by what
factor is the heavier person’s height greater than the
shorter person’s height?
--
Let the 2nd person's weight be "x"
x = k*H^3
H = [x/k]^(1/3) (2nd person's height)
-------------------
The 1st person's weight is "2x"
2x = kH^3
H = [2x/k]^(1/3)
H = 2^(1/3)[x/k]^(1/3) (1st person's height)
-----
By what factor is the heavier person’s height greater than the
shorter person’s height?
Answer: 2^(1/3)
=========================
Cheers,
Stan H.
================
|
Word_Problems_With_Coins/424032: Jeffery has a piggy bank with $28.00 in dimes and quarters. If there are 205 coins in total, how many of each coin does he have? 1 solutions
Answer 295703 by stanbon(57307) on 2011-03-18 22:59:56 (Show Source):
You can put this solution on YOUR website!Jeffery has a piggy bank with $28.00 in dimes and quarters. If there are 205 coins in total, how many of each coin does he have?
-------------------
Equations:
Quantity Eq.::: d + q = 205 coins
Value Eq.:::: 10d + 25q = 2800 cents
-------------------------------------------
Multiply thru the Quantity Eq. by 10 to get:
10d + 10q = 2050
10d + 25q = 2800
----------------------
Subtract and solve for "q":
15q = 750
q = 50 (# of quarters)
---
Solve for "d":
d + q = 205
d + 50 = 205
d = 155 (# of dimes)
======================
Cheers,
Stan H.
===============
|
Polynomials-and-rational-expressions/424026: A little help please?!...
Suppose that the price, p, and the quantity demanded, q, are related to the equation:
p=200-5q
Find the price if the quantity demanded is 15 units.
Thank you!! :) 1 solutions
Answer 295702 by stanbon(57307) on 2011-03-18 22:54:17 (Show Source):
You can put this solution on YOUR website!Suppose that the price, p, and the quantity demanded, q, are related to the equation:
p=200-5q
Find the price if the quantity demanded is 15 units.
----
price = 200-5*15 = 200-75 = 125
========================================
Cheers,
Stan H.
===========
|
Permutations/424027: How many ways can a 4 person committee be formed from a group of 8 people, if Jon must be on the committee. (Jon is part of the group of 8 people) 1 solutions
Answer 295698 by stanbon(57307) on 2011-03-18 22:51:44 (Show Source):
You can put this solution on YOUR website!How many ways can a 4 person committee be formed from a group of 8 people, if Jon must be on the committee. (Jon is part of the group of 8 people)
-----------------------------
If Jon is on the committee, a group of 7 remains:
7C3 = (7*6*5)/(1*2*3) = 42
=================================
Cheers,
Stan H.
|
Radicals/424029: Find a number such that twice its square root is 36. 1 solutions
Answer 295696 by stanbon(57307) on 2011-03-18 22:44:46 (Show Source):
You can put this solution on YOUR website!Find a number such that twice its square root is 36.
---
Let the number be "x":
Equation:
x^2=36
---
x^2-36 = 0
(x-6)(x+6) = 0
---
x = 6 or x = -6
=======================
cheers,
Stan H.
=================
|
test/424031: Frank is two thirds as a Adam. In seven years, frank will be three fourths as old as adam.how old are they now ? 1 solutions
Answer 295694 by stanbon(57307) on 2011-03-18 22:42:31 (Show Source):
You can put this solution on YOUR website! Frank is two thirds as a Adam. In seven years, frank will be three fourths as old as adam.how old are they now ?
-----------------------
Equations:
f = (2/3)a
(f+7) = (3/4)(a+7)
------
Substitute for "f" and solve for "a":
------
((2/3)a + 7) = (3/4)a + (21/4)
----
(2/3)a - (3/4)a = (21/4)-7
(-1/12)a = -7/4
a = 21 yrs old (Adam's age now)
-------
Solve for "f":
f = (2/3)a
f = (2/3)21
f = 14 yrs old (Frank's age now)
===================================
Cheers,
Stan H.
======================
|
Graphs/424028: A little help please!?...
Find the two x-intercepts of the graph of:
g(x)=2x^2-8x+6
*Use the answer format of "x=#, x=#"
Thank you in advance! :) 1 solutions
Answer 295693 by stanbon(57307) on 2011-03-18 22:35:43 (Show Source):
You can put this solution on YOUR website!Find the two x-intercepts of the graph of:
g(x)=2x^2-8x+6
---
Solve 2x^2-8x+6 = 0
x^2-4x+3 = 0
(x-3)(x-1) = 0
----
x-intercepts are x = 3 and x = 1
===================================
Cheers,
Stan H.
|
Probability-and-statistics/423732: From a group of 7 men and 5 women...
a. How many committees of 4 can be formed?
b. How many ways can a President, vice president and secretary be chosen?
c. How many committees of 2 women and 3 men can be formed? 1 solutions
Answer 295681 by stanbon(57307) on 2011-03-18 21:46:14 (Show Source):
You can put this solution on YOUR website!From a group of 7 men and 5 women...
a. How many committees of 4 can be formed?::::12C4 = 495
-------------------------------
b. How many ways can a President, vice president and secretary be chosen?
Ans: 12*11*10 = 1320
-----------------------
c. How many committees of 2 women and 3 men can be formed?
Ans: 5C2*7C3 = 350
========================
Cheers,
Stan H.
===============
|
Probability-and-statistics/424012: State College is evaluating a new English composition course for freshmen. A random sample of n=25 freshmen is obtained and the students are placed in the course during their first semester. One year later, a writing sample is obtained for each student and the writing samples are graded using a standardized evaluation technique. The average score for the sample is M=76. For the general population of college students, writing scores form a normal distribution with a mean of m=70.
a. If the writing scores for the population have a standard deviation of σ=20, does the sample provide enough evidence to conclude that the new composition course has a significant effect? Assume a two- tailed test with α=.05.
1 solutions
Answer 295679 by stanbon(57307) on 2011-03-18 21:38:46 (Show Source):
You can put this solution on YOUR website!State College is evaluating a new English composition course for freshmen.
-------------------------------
A random sample of n=25 freshmen is obtained and the students are placed in the course during their first semester. One year later, a writing sample is obtained for each student and the writing samples are graded using a standardized evaluation technique.
--------------------------------------
The average score for the sample is M=76. For the general population of college students, writing scores form a normal distribution with a mean of m=70.
---------------------------
a. If the writing scores for the population have a standard deviation of σ=20, does the sample provide enough evidence to conclude that the new composition course has a significant effect? Assume a two- tailed test with α=.05.
---
Ho: u = 70
Ha: u > 70
-------------
Note: You are running sample results for a mean against population
results for a mean. That is not a 2-tail test; it is a right-
tail test.
-----
I ran a T-test on the data and got the following:
test statistic: t = 1.5
p-value = 0.0733
-----------------------
Conclusion: Since the p-value is greater than 5%, fail to
reject Ho.
==============
Cheers,
Stan H.
===========
|
Polynomials-and-rational-expressions/423992: A launched rocket has an altitude in meters, given by the polynomial h+vt-4.9^2, h is the height in meters v is the velocity in meters per second and t is the number of seconds for which it takes the rocket to become airborne. If the rocket is launched from the top of a tower that is 120 meters and the initial speed is 40 meters per second, what will its height be after 3 seconds??
the height will be __meters? 1 solutions
Answer 295658 by stanbon(57307) on 2011-03-18 20:16:26 (Show Source):
You can put this solution on YOUR website! A launched rocket has an altitude in meters, given by the polynomial
h(t)= -4.9t^2 + vot + so, h is the height in meters vo is the velocity in meters per second, so is the initial height of the object and t is the number of seconds the rocket is airborne.
If the rocket is launched from the top of a tower that is 120 meters and the initial speed is 40 meters per second, what will its height be after 3 seconds??
--------
h(3) = -4.9(3)^2+40(3) + 120
-----------------
h(3) = 195.9 meters
============================
Cheers,
Stan H.
|
Graphs/423997: graph the line containing the given pair of points and find the slope (2,0) and (8,-1) 1 solutions
Answer 295654 by stanbon(57307) on 2011-03-18 20:06:30 (Show Source):
You can put this solution on YOUR website!graph the line containing the given pair
of points and find the slope (2,0) and (8,-1)
===============
slope = (-1-0)/(8-2) = -1/6
-----
intercept:
0 = (-1/6)2 + b
----
b = 1/3
------
Equation:
y = (-1/6)x+ (1/3)
========================

Cheers,
Stan H.
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Quadratic_Equations/423981: Consider the quadratic function; f(x)=r^2 + (x+s)(x-s) here r,s are arbitary constants, find the min possible value for f and also the value of x for which the min is achieved.
my email address is jesssie87@yahoo.com.au 1 solutions
Answer 295652 by stanbon(57307) on 2011-03-18 20:02:25 (Show Source):
You can put this solution on YOUR website!Consider the quadratic function; f(x)=r^2 + (x+s)(x-s) here r,s are arbitary constants, find the min possible value for f and also the value of x for which the min is achieved.
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f(x)=r^2 + (x+s)(x-s)
f(x) = x^2-s^2+r^2
a = 1 ; b = 0 ; c = (r^2-s^2)
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The minimum value occurs when x = -b/(2a) = 0/(2) = 0
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f(0) = r^2-s^2
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Min: (0, r^2-s^2)
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Cheers,
Stan H.
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Age_Word_Problems/423995: Al is five times as old as Tim. In five years,Al will be four times as old as Tim. How old will Tim be in five years? 1 solutions
Answer 295649 by stanbon(57307) on 2011-03-18 19:57:58 (Show Source):
You can put this solution on YOUR website!Al is five times as old as Tim. In five years,Al will be four times as old as Tim. How old will Tim be in five years?
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Equations:
a = 5t
(a+5) = 4(t+5)
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Substitute for "a" and solve for "t":
(5t+5) = 4t+20
t = 15 (Tim's age now)
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How old will Tim be in five years?
t+5 = 20 yrs old
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Cheers,
Stan H.
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Quadratic_Equations/423723: Help me to solve:
10x^4 - 19x^2 + 6 = 0 1 solutions
Answer 295444 by stanbon(57307) on 2011-03-17 22:18:45 (Show Source):
You can put this solution on YOUR website!10x^4 - 19x^2 + 6 = 0
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10x^4-15x^2-4x^2+6 = 0
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5x^2(2x^2-3)-2(2x^2-3) = 0
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(2x^2-3)(5x^2-2) = 0
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Solve:
2x^2-3 = 0
x^2 = 3/2
x = +-sqrt(6)/2
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Solve:
5x^2-2 = 0
x^2 = 2/5
x = +-sqrt(10)/5
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Cheers,
Stan H.
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Probability-and-statistics/423725: How many different arrangements of three coins can be made if you have a penny, a nickel, a dime, a quarter, and a silver dollar? 1 solutions
Answer 295443 by stanbon(57307) on 2011-03-17 22:10:34 (Show Source):
You can put this solution on YOUR website!How many different arrangements of three coins can be made if you have a penny, a nickel, a dime, a quarter, and a silver dollar?
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Ans: 5C3 = 5C2 = 10
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Cheers,
Stan H.
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Probability-and-statistics/423697: ACCORDING TO A RECENT STUDY 38% OF ALL WOMEN WILL SUFFER A HIP FRACTURE BECAUSE OF OSTEOPOROSIS BY THE AGE OF 85. IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT
A) NONE OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
b) AT LEAST 4 OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
c)FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS 1 solutions
Answer 295442 by stanbon(57307) on 2011-03-17 22:09:02 (Show Source):
You can put this solution on YOUR website!ACCORDING TO A RECENT STUDY 38% OF ALL WOMEN WILL SUFFER A HIP FRACTURE BECAUSE OF OSTEOPOROSIS BY THE AGE OF 85.
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P(will suffer) = 0.38
P(will not suffer) = 0.62
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IF SIX WOMEN AGED 85 ARE RANDOMLY SELECTED WHAT IS THE PROBABILITY THAT
A) NONE OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
P(x = 0) = (0.62)^6
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b) AT LEAST 4 OF THEM WILL SUFFER (OR HAS SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
P(4<= x <= 6) = 1 - P(x<= 0 <=3) = 1-binomcdf(6,0.38,3) = 0.1527
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c)FEWER THAN TWO OF THEM WILL SUFFER (OR HAVE SUFFERED) A HIP FRACTURE DUE TO OSTEOPOROSIS
P(0<= x <=2) = binomcdf(6,0.38,2) = 0.5857
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Cheers,
Stan H.
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Probability-and-statistics/423704: 70% OF THE STUDENT BODY OF A very large post secondary institution is female. in a random sample of 12 students what is the probability that
a)at most half will be females?
b) more than 7 will be femels?
c) fewer than 3 will be males 1 solutions
Answer 295435 by stanbon(57307) on 2011-03-17 21:38:17 (Show Source):
You can put this solution on YOUR website!70% OF THE STUDENT BODY OF A very large post secondary institution is female. in a random sample of 12 students what is the probability that
a)at most half will be females?
P(0<= x <=6) = 12C6(0.7)^6*(0.3)^6 = binompdf(12,0.7.6) = 0.0792
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b) more than 7 will be females?
P(7<= x <=12) = 1-answer to part a = 1-0.0792 = 0.9208
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c) fewer than 3 will be males
P(0<= x <=2) = 12C3(0.3)^3*(0.7)^9 = 0.1678
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Cheers,
Stan H.
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Rational-functions/423694: I need help!
Write a monomial and a polynomial using x as the variable. Find their product.
Thank you 1 solutions
Answer 295429 by stanbon(57307) on 2011-03-17 21:25:17 (Show Source):
You can put this solution on YOUR website!Write a monomial and a polynomial using x as the variable. Find their product.
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Monomial: 3x
Polynomial: x^2+2x+1
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Product: 3x(x^2+2x+1) = 3x^3+6x^2+3x
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cheers,
Stan H.
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Angles/423666: find the measures of each angle of the triangle and then give the order of the lengths of the sides from least to greastest.
after you have written the sentence for each step,illustrate the step with the information given in the example .include a sketch of the triangle labeled with the pertinent information
example:the angles of triangle ABC,in order,are in the ratio 5:7:8 1 solutions
Answer 295414 by stanbon(57307) on 2011-03-17 20:36:36 (Show Source):
You can put this solution on YOUR website!find the measures of each angle of the triangle and then give the order of the lengths of the sides from least to greastest.
after you have written the sentence for each step,illustrate the step with the information given in the example .include a sketch of the triangle labeled with the pertinent information
example:the angles of triangle ABC,in order,are in the ratio 5:7:8
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smallest: 5x
middle: 7x
largest: 8x
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Equation:
5x+7x+8x = 180
20x = 180
x = 9
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smallest: 5x = 45 degrees
middle: 7x = 63 degrees
largest: 8x = 72 degrees
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Cheers,
Stan H.
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Probability-and-statistics/423549: Show all work. Mary purchased a package of 13 different plants, but she only needed 9 plants for planting. In how many ways can she select the 9 plants from the package to be planted? 1 solutions
Answer 295411 by stanbon(57307) on 2011-03-17 20:16:21 (Show Source):
You can put this solution on YOUR website!Mary purchased a package of 13 different plants, but she only needed 9 plants for planting. In how many ways can she select the 9 plants from the package to be planted?
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13C9 = 13C4 = (13*12*11*10)/(1*2*3*4) =715
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Cheers,
Stan H.
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