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69% from program X enroll in college,46.2% from program Y: What %-age more from prog. X attend clg?

I need to know the formula for figuring out this type of comparison: 69% of students from program X enroll in college and 46.2% of students from program Y enroll in college; please help me fill in the blank: ___ % more participants from program X attend college than program Y.

Research, Knowledge and Information :


stics of Un e Students 1 th and Verbal SAT scores (senior ...


ents enroll Percent nrolled in On medial Cours 69.7 ... 46.2 53.8 00.0 tional ... The prog students who gr their class and college. 200 9 deral, ...
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The Condition of Education - Participation in Education ...


Between 2000 and 2015, total undergraduate enrollment in degree-granting postsecondary institutions increased by 30 percent (from 13.2 million to 17.0 million).
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DRAFT – March 18, 2011 - Shoreline Community College


46. Program Strengths ... most students surveyed indicated they would not enroll in a VCT program taught ... If the college cannot add more computer ...
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Fiscal Year 2009-2010 - New Jersey


unity College October z PO Box 330 2010 z Sourc ... 2 3 4 Number o. Undergr. Number . Number. Undupli ... ls by Attend ire academ
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Medical Services Standards of Medical Fitness


Medical Services Standards of Medical Fitness ... (2) Pure tone level not more than 45 dB at 3000 cycles per second or 55 dB at 4000 ... 2 Maximum weight by years of age
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1. Introduction


Age (years), mean (SD) 46.3 ... in addition to asking the program participants to attend six ... vs. attending a free program: Hold more community education ...
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www.rowan.edu


o Resource persons may be invited to attend meetings of the ... PROGRAM. The College offers its full ... Leave Donation Program, the employee can enroll or re ...
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Make a Refundable deposite



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Express HelpLine


Express Helpline- Get answer of your question fast from real experts.
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Suggested Questions And Answer :


69% from program X enroll in college,46.2% from program Y: What %-age more from prog. X attend clg?

???????????????? "atnd sig" ?????????? ?? maebee yu wanna no how many in class?????? ??? Iz this won av em liberal komplantes that 70% av wites gotu Kolleg, but oenlee 8% av blaks???? Hard tu get NE numbers kauz yu dont suppli NE numbers
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1) If you enrolled in a 128 hour program, how many semesters would it take you to complete your degree taking 8 credit hours a semester?

128/8=16 semesters
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The interest due is ​$ round to the nearest cent

4 months is 1/3 of a year. Assuming simple interest at 6.9% per year, 1/3 would be 6.9/3=2.3%. And 2.3*19000/100=$437.
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If you have 10 boys and 20 girls how would you make a comparison using fractions?

10/30 are boys aka 1/3 20/30 are girls aka 2/3 there ya go
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When Ann is twice as old as Ruth was when Ann was half Ruth's present age,

Ann is currently 44 years old and Ruth is 54. How did I get this? R=Ruth's present age and A=Ann's present age. There seemto be 4 time zones: y years ago, now, x years hence, z years hence. Let's start in the past. A-y=R/2. y years ago Ann was half Ruth's present age. At that time Ruth was R-y years old. Now the future. When Ann is twice as old as this in x years' time we have A+x=2(R-y). At the same time Ruth's age will be half as much again (3/2) as Ann's will be in z year's time: R+x=3/2(A+z). In z year's time Ruth will be R+z and Ann will be A+z years old, and Ruth will be half again as old as Ann is now: R+z=3/2A.  We can start to eliminate some variables. y=A-R/2 from the first equation. So A+x=2(R-A+R/2) substituting in the second equation. A+x=3R-2A. z=3/2A-R from last equation is last paragraph. So R+x=3/2(A+3/2A-R)=3/2(5/2A-R). We now have two simultaneous equations from which we can eliminate x: A-R=3R-2A-3/2(5/2A-R). A-R=9/2R-23A/4, so, getting rid of the fractions, 4A-4R=18R-23A. 27A=22R, from which A=22R/27 or R=27A/22. We know that either Ruth or Ann is aged between 50 and 59. We also know that all ages are whole numbers. The only multiple of either 22 or 27 between 50 and 59 is 54=2*27. So R=54 and A=44. Therefore Ruth is 54 and Ann is 44. The original question can now be read knowing the women's ages to confirm their correctness. In the course of doing so, x, y and z will be established. y=17, z=12 and x=30 years. See table below where age relationships should look clearer. Every age is related to another across time. For example, Ruth's age in 12 years' time is half as much again as Ann's present age, and Ruth's age in 30 years' time is half as much again as Ann's in 12 years' time.               -17   Now +12 +30        Ruth  37    54     66    84            Ann  27    44     56    74    
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Word Problem where you have to figure out Goerge's wife's age in 1941.

In 1941 George's children were both between the ages of 10 and 20. The sum of the cube of one child's age and the square of the other child's age gives the year in which George's wife was born. How old was his wife in 1941? Start by looking for the possible cubed age: 10^3 = 1000 11^3 = 1331 12^3 = 1728 13^3 = 2197 We can stop there. Obviously the child's age that is cubed can't be 13; 2197 is nearly 200 years in the future. Now, look for the age that is squared. Start at the top, 20. 20^2 = 400 Add 400 to 10^3: we get 1400, too far in the past. Add 400 to 11^3: we get 1731, also too far in the past. Add 400 to 12^3: we get 2128, at least a hundred years in the future. We have narrowed the cubed age to 12. Now we need a squared age that gives a reasonable sum for the mother's birth year. 10^2 = 100: add to 12^3: we get 1828, too far in the past. 11^2 = 121: add to 12^3: we get 1849, still too far in the past. 12^2 = 144: add to 12^3: we get 1872, too far back. 13^2 = 169: add to 12^3: we get 1897; age in 1941 = 44; possible. 14^2 = 196: add to 12^3: we get 1924; age in 1941 = 17; impossible. George's wife was born in 1897, and she was 44 in 1941. One child was 12, the other was 13.
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The sum of the ages of two brothers is 38. Four years ago the elder brothers age was square that of the younger brother. What are the ages of the two brothers?

The sum of the ages of two brothers is 38. Four years ago the age of the elder brother was square that of the younger brother. What are the ages of the two brothers? 4 years ago, each brother would have been 4 years younger. That means that the sum of their ages would have been 8 less than the sum of therir ages now. (8 = 2 * 4) So, 8 years ago the sum of their ages would gave been 30. The only two numbers, where one of them is the square of the other, and fit in this range, are 5 and 25. The ages of the brothers are: 9 yrs and 29 yrs
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jans age is 3 times pams age sue is twice as old as pam the sum of their age is 30 how old is each girl

jans age is 3 times pams age sue is twice as old as pam the sum of their age is 30 how old is each girl i need to know how old each girl is  jan is 3 times pams age and sue is twice as old as pam the sum of their ages is 30   j = jan s = sue p = pam   j = 3p call this eq1 s=2p ← eq2 j + p + s = 30 ← eq3 substitute eq1,eq2 to eq3   3p + p + 2p = 30 6p = 30   pam age = 5   in eq 1 j = 3p j = 3(5)   j an age = 15   in eq2 s = 2p s = 2(5)   Sue age =10
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The age of mrs. Laurel is 9 years more than 3 times her aon's age. Find the age of each if fifteen years from now she will be 6 years more than twice her son's age.

The age of mrs. Laurel is 9 years more than 3 times her son's age. Find the age of each if fifteen years from now she will be 6 years more than twice her son's age. Let L = mrs Laurel's current age Let S = the son's current age L = 3S + 9        (mrs. Laurel is 9 years more than 3 times her son's age ​L + 15 = 2(S + 15) + 6       (fifteen years from now she will be 6 years more than twice her son's age. Substituting for L = 3S + 9 into teh 2nd equation, 3S + 9 + 15 = 2S + 30 + 6 S + 24 = 36 S = 12 L = 3S + 9 = 36 + 9 = 45 The ages are: Mum = 45, Son = 12
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Age Problem

Let Mark's age be M. Let the age of his son be B (boy) now. 3 years ago his son was B-3 years old, and in 3 years' time Mark will be M+3 years old, so M+3=9(B-3); M+3=9B-27, M=9B-30. Let the youngest girl's age be Y, so B=4Y.  If we put the children's ages in order we have a number of possibilities: Y G G G 4Y Y G G 4Y G Y G 4Y G G Y 4Y G G G One of the G's (girls) is 7Y and therefore older than B=4Y. So the boy can't be the eldest. So we eliminate (1). What about (2)? The ages are 3 years apart so if the eldest G=7Y then 7Y-4Y=3 and 3Y=3 making Y=1, the age of the youngest girl. That makes the ages 1 4 7 10 14 which actually fits (4). So the boy is 4, the youngest is 1 and the middle girls is 7. M=9B-30=6 which cannot be! That seems to eliminate (2). Now (3): 7Y-4Y=6, 3Y=6, Y=2: the ages are 2 5 8 11 14, the boy is 8 and the youngest is 2. M=9B-30=42. That seems reasonable: Mark is 42, his son is 8 and his 4 daughters are 2, 5, 11 and 14.  
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