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Calculate Table volume at 5 years using the following information. Field Avg. size (characters) Student ID 8 Student name 25 Phone no 11 Admission date 7 Overhead 25% Initial table size 1000 Initial table volume 23000 Growth rate/month 300 

          Calculate Table volume at 5 years using the following information.
Field
Avg. size (characters)
Student ID
8
Student name
25
Phone no
11
Admission date
7
Overhead
25%
Initial table size
1000
Initial table volume
23000
Growth rate/month
300
Show more…
Calculate Table volume at 5 years using the following information. Field Avg. size (characters) Student ID 8 Student name 25 Phone no 11 Admission date 7 Overhead 25% Initial table size 1000 Initial table volume 23000 Growth rate/month 300

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Computer Science and Information Technology
Computer Science and Information Technology
Trishna Knowledge Systems 2018 Edition
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Calculate the volume of the table after 5 years using the following information: Field Avg. size (characters) Student ID Student name Phone no Admission date 8 25 11 7 Overhead 25% Initial table size Initial table volume 1000 23000 Growth rate/month 300
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Transcript

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00:01 The concept involved in this problem is to model a situation with a quadratic function.
00:08 In this problem, you're given a table that lists the volume of non -federal student loans in billions of dollars represented in terms of the number of years since 2004.
00:22 So in 2004, there were $15 .1 billion in non -federal students.
00:31 Loans.
00:31 So that order to repair would be 015 .1.
00:38 Okay, then in 2006, which x would have been two, there were 20 .5 billion dollars of student loans.
00:52 And then in 2008, which x would have been four, there were 11 .0 billion dollars in non -federal student loans.
01:02 Okay, we are to use this data and write a quadratic function that will fit this data.
01:10 So using the standard form for a quadratic function, i'm going to use each of the three -ordered pairs to write an equation.
01:18 So i'm going to start off with 15 .1 is equal to a times 0 squared plus b times 0 plus c, which will get us.
01:35 That c equals 15 .1.
01:39 Second order pair 20 .5 is equal to a times 2 squared plus b times 2 plus c.
01:51 Okay, this will get the equation 4a plus 2b plus c is equal to 20 and then the third order pair 11 .0 is equal to a times 4 squared plus b times 4 plus c and that will get us the equation 16a plus 4b plus c is equal to 11 .0.
02:26 So now what i'm going to do is i'm going to take each of this, i'm going to take the second and the third equation and i'm going to replace the c in each one of those with the 15 .1 that we found c to b.
02:44 Okay, so using the 4a plus 2b plus 15 .1 is equal to 20 .5.
02:56 Okay, if i subtract 15 .1 from each side, 4a plus 2b is equal to 5 .4.
03:06 And i'm going to reduce that equation by dividing each term by 2.
03:12 So that would be 2a plus b.
03:16 Is equal to 2 .7.
03:19 Okay, then i'm going to work in a similar way with that third equation.
03:24 Replacing c with 15 .1.
03:27 So there's 16a plus 4b plus 15 .1 is equal to 11 .0.
03:38 Okay, subtracting 15 .1 from each side, 16a plus 4b is equal to a negative 4 .1...
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