12.Pt= 75000(1+0.025)^-15 ANSWER=51,784.91676 or 51,785 people 13. 1,000,000(1-0.10)^7=478296.9 ANSWER= 7 years 14. Nt=100(1+0.05)^33=500.32 ANSWER= 33 days
HOMEWORK YESTERDAY; September 12, 2012 Math Homework p. 355-356 numbers 1, 2, 3, 4, 6, 8, 9, 12, 13 and 14.
The ones with ✓ are my final answers. :)
Solutions: Problem #1 Pt = P(1+r)^n 4 years = 12 months/1 year (cancel out years, month was left) so, 48 months/6 = 8 months Pt = 50,000(1+.08)^8 = 50,000(1.85) = 92,500 people ✓
Problem #2 Nt = No(3)^x/k = 100(3)^6/1 = 72,900 bacteria ✓
Problem #3 P = 5000e^(.15y) y = 2008-2000 = 8(.15) = 1.2 P = 5,000e^1.2 = 16,600.58 people ✓
Problem #4 Nt = No(1/4)^x/k = 50(1/4)^4/1 = 0.2 grams ✓
Problem #6 Nt = No(1/2)^x/k = 100(1/2)^168/12 = 100 (1/2)^14 = 0.006 grams ✓
Problem #8 N = 8(2)^t 452 = 8(2)^t (divide both sides by 8) 56.5 = (2)^t log56.5 = log2 = 5.82 hours ✓
Problem #9 Use the formula given in number 8 which is N=8(2)^t N = 8(2)^10 = 8,192 bacteria ✓
Problem #12 Pt = P(1+r)^n = 75,000 (1+.025)^-15 = 75,000 (1.025)^-15 = 51,784 people ✓
Problem #13 Q = p(1-d)^n 500,000 = 1,000,000(.9)^n (divide both sides by 1,000,000) so, .5=(.9)^n log.5/log.9 = 7 years ✓
Problem #14 Pt = p(1+r)^n 500 = 100(1+.05)^n (divide both sidea by 100) 5 = (1.05)^n log5/log1.05 = 33 days ✓
1.) Pn= P(1+r)^n = 50,000(1+0.08)^4 P4= 68,024 people There are 68,024 people after 4 years.
2.) Nt = No (2)^x/k = 100 bacteria (3)^6 Nt= 72,900 bacteria There will be 72,900 bacteria in the petri dish after 6 hours.
3.) Pn= 5000(2.718)^(0.15x8) y= 2008-2000 y= 8 P8= 16601 people The human population will be 16,601 at the year 2008
4.) Nt= No (1/4)^x/k = 500g (1/4)^4 Nt= 2g The substance weight would be 2g after 4 minutes.
5.) Nt= No (1/2)^x/k = 100g(1/2)^14 Nt= 0.0061g After one week, the remaining substance would be 0.0061g.
6.) N= 8(2)^t 452= 8(2)^t 452/8= (2)^t Log(452/8)= t [log(2)] T= 5.82 hours The culture needs 5.82 hours to grow 452 bacteria.
7.) N= 8(2)^10 = 8,192 bacteria There will be 8,192 bacteria after 10 hours.
8.) Q= 75000 (1+.025)^-15 = 75000 (.6904) = 51,784.9 or 51,785 The human population of the province 15 years ago was 51,785.
9.) Q = P(1-d)^t 500000= 1000000(1-0.10)^t 500000/1000000= (1-0.10)^t log (500000/1000000)= t [log(1-0.10)] t= log (500,000/1,000,000) / [log(1-0.10)] t= 1.4 years It would take 1.4 years buy the car at half its price.
10.) Pn = P(1+r)n 500 = 100(1+0.05)^t 500/100 = (1+0.05)^t log (500/100) = t [log(1+0.05)] t= log (500/100)/[log(1+ 0.05)] t = 32. 97 or 33 days It would take the SARS virus 33 approximately 33 days to infect 500 people in China.
14. Pn = P(1+r)^n 500 = 100(1+0.05)^t 500/100 = (1+0.05)^t log (500/100) = t [log(1+0.05)] t= log (500/100) / [log(1+ 0.05)] t = 32. 97 days t= 33 days written math (page355) 5) A = 8,000(1-.10/12)^12(1) A = P 7,235.67
7) A = 10,000(1+.04/12)^12(2) A = P 10,831.43
10) A = 100,000(1+.09/4)^4(5) A = P 156,050.92
13) 500,000 = 1,000,000(1-0.10)^t 500,000/1,000,000 = (.9)^t t = log (1/2) / [log(.9)] t = 6.58 years t = 7 years
15) Banko Matatag: A = 1,000,000(1+.06/2)^2(5) A = P 1,343,916.38
Banko Subok: A = 1,000,000(1+.05/4)^4(5) A = P 1,282,037.23 I would like to deposit my money to banko matatag because i can earn more haha
15) A=P (1+r/n)^n Banko matatag 1.000.000 (1+.06/2)^(2)(5) = P 1.343.916.38 Banko subok 1,000,000 (1+0.5/4)^(4)(5) P = 1,282,037.23 MY CHOICE IS BANKO MATATAG
MATH AND I Throughout the high school years, every student has their own opinions about the subjects they like and dislike. This does not mean the literal meaning. It means that it is about their strengths and weaknesses when it comes to academics. These opinions often change throughout the time. There are certain standards in every individual in order to succeed this kind of high school life. “Mathematics is an essential part of every high school student’s education.” This quote says it all. It explains the subject Math in eleven words. Every student learns certain lessons in math that will carry with them through life. Each of these are building blocks to help shape who we become. As a math student, I feel math has taught me not only about addition, subtraction, division... But it has taught me thinking skills and problem solving that will take me further than just math class. There are times in which you have to decide which path you will take after high school, although every path is different and you will take the skills in which you have learned in math class to a different level. When I was young, I used to say “I hate math”, although this sentence did not last long. As I was growing up, I slowly understood what math really meant and how we can apply it in real life situations. I used to say that this has nothing to do with shopping or dancing and such and that I won’t need this in the future. For my other batchmates, these are hard, most especially when you are surrounded by your best friends. Concentration is obviously needed when it comes to this matter. The dedication that this class requires teaches every student concentration. The concentration on this one course will help high school students pursue their dreams while taking other courses to achieve this. Another major influence on a high school student’s math education is the teacher. A teacher is crucial to a student’s success. The ideal teacher knows several ways to explain a certain concept. Doc Ed is an example of an ideal teacher. Soon enough, I know that I will become successful because of math. I know that this will be a big help in my future. I just have to focus at this moment and I can’t be distracted by the things that will destroy my future.
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ReplyDelete1.P = 50,000 (1+ 0.08)^8/0.5
ReplyDeleteANSWER=17,297.13 or 17,297 people
2. Nt = 100(3)^6
ANSWER=72,900 bacteria
3.P= (5000)(2.71828)^(0.15)(8)
ANSWER= 16,600.57 or 16,601 people
4.Nt= 50(0.25)^4
ANSWER=0.1953125 grams or 0.20 grams
6.Nt=100(1/2)^14
6.1035x10^-3 or 0.006 bacteria
8
452=8(2)^t
452/8=2^t
log (452/8)=t[log(2)]
t= log (452/8)/log(2)
t= 5.82 hours
9.Nt=8(2)^10
ANSWER=8,192 bacteria
12.Pt= 75000(1+0.025)^-15
ANSWER=51,784.91676 or 51,785 people
13. 1,000,000(1-0.10)^7=478296.9
ANSWER= 7 years
14. Nt=100(1+0.05)^33=500.32
ANSWER= 33 days
100%
DeleteVery good
Fristine Chua
ReplyDeleteBk. Page 355
1. P= 50,000
r= 0.08
n= 4
initial time= 6 or 0.5
P4 = 50,000 (1+ 0.08) 8/0.5
17,297.13
2. 100 bacteria
3x every 60 mins
6 hours
P6 = 100(3)6
72,900
3. P= (5000)(2.71828)(0.15)(8)
16,600.57 or 16,601
4. P= 50(0.25)4
0.1953125 or 0
6. P=100(1/2)14
6.1035x10-3 or 6
8-9. 8(2)t
8 N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 hours
N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
12. 75000(1+0.025)-15
51,784.91676 or 51,785
13. 1,000,000(1-0.10)something
1,000,000(0.90)7
478296.9
7 years
14. 100(1+0.05)33
500.32 answer : 33 days
100
DeleteVery good
1. P=50,000 [1+.08]^12*4/6
Delete50,000 [ 1.08 ]^8
50,000 [1.85]
=92,500 people
2. N=100 [3]^6
=72,900 bacteria
3. P=5,000e^[.15*8]
P=5,000e^1.2
P=16,600 people
4. N=50[1/4]^4
=0.2 grams
6. N=100 [1/2]^14
=0.006 grams
8. 452=8[2]^t
log base 2 of 56.5=t
t=5.82 hrs
9. N=8[2]^10
=8 [1024]
=8,192 bacteria
12. 75,000 [1.025]^-15
51,784 people
13. 500,000=1,000,000 [1-0.1]^n
.5=.9^n
about 7 years
14. 500=100 [1+.05]^n
5=1.05^n
33 days
1. P=50,000 [1+.08]^12*4/6
Delete50,000 [ 1.08 ]^8
50,000 [1.85]
=92,500 people
2. N=100 [3]^6
=72,900 bacteria
3. P=5,000e^[.15*8]
P=5,000e^1.2
P=16,600 people
4. N=50[1/4]^4
=0.2 grams
6. N=100 [1/2]^14
=0.006 grams
8. 452=8[2]^t
log base 2 of 56.5=t
t=5.82 hrs
9. N=8[2]^10
=8 [1024]
=8,192 bacteria
12. 75,000 [1.025]^-15
51,784 people
13. 500,000=1,000,000 [1-0.1]^n
.5=.9^n
about 7 years
14. 500=100 [1+.05]^n
5=1.05^n
33 days
Short solution
Delete5. A=8000 [ 1-.1/12 ]^12
ReplyDelete=P 7,235.67
7. A=10000 [1+.04/12]^24
=P 10,831.43
10. A=100000 [1+.09/4]^20
=P 156,050.92
13. 500000=1000000 [1-.1]^n
1/2=.9^n
log.5/log.9=n
n=6.58
15. FOR BANKO MATATAG:
A=1000000[1+.06/2]^10
=P 1,343,916.38
FOR BANKO SUBOK
A=1000000[1+.05/4]^20
=P 1,282,037.23
I would deposit my money in BANKO MATATAG.
Incomplete
DeleteHOMEWORK YESTERDAY; September 12, 2012
ReplyDeleteMath Homework p. 355-356 numbers 1, 2, 3, 4, 6, 8, 9, 12, 13 and 14.
The ones with ✓ are my final answers. :)
Solutions:
Problem #1
Pt = P(1+r)^n
4 years = 12 months/1 year (cancel out years, month was left) so, 48 months/6
= 8 months
Pt = 50,000(1+.08)^8
= 50,000(1.85)
= 92,500 people ✓
Problem #2
Nt = No(3)^x/k
= 100(3)^6/1
= 72,900 bacteria ✓
Problem #3
P = 5000e^(.15y)
y = 2008-2000
= 8(.15)
= 1.2
P = 5,000e^1.2
= 16,600.58 people ✓
Problem #4
Nt = No(1/4)^x/k
= 50(1/4)^4/1
= 0.2 grams ✓
Problem #6
Nt = No(1/2)^x/k
= 100(1/2)^168/12
= 100 (1/2)^14
= 0.006 grams ✓
Problem #8
N = 8(2)^t
452 = 8(2)^t (divide both sides by 8)
56.5 = (2)^t
log56.5 = log2
= 5.82 hours ✓
Problem #9
Use the formula given in number 8 which is N=8(2)^t
N = 8(2)^10
= 8,192 bacteria ✓
Problem #12
Pt = P(1+r)^n
= 75,000 (1+.025)^-15
= 75,000 (1.025)^-15
= 51,784 people ✓
Problem #13
Q = p(1-d)^n
500,000 = 1,000,000(.9)^n (divide both sides by 1,000,000)
so, .5=(.9)^n
log.5/log.9
= 7 years ✓
Problem #14
Pt = p(1+r)^n
500 = 100(1+.05)^n (divide both sidea by 100)
5 = (1.05)^n
log5/log1.05
= 33 days ✓
100
DeleteVery very good

ReplyDeleteAlexandra Cuartero
Written Math (p.355-356)
1. Pn = P(1+r)^n
= 50,000 (1+0.08)^4
P4 = 68,024 people
2. Nt = No (2)^x/k
= 100 bacteria (3)^6/1
Nt = 72,900 bacteria
3. Pn = 5000e^(0.15*8)
= 5000e^1.2 ≈ 5000(3.320116923 )
P8 = 16601 people
4. Nt = No (1/2)^x/k
= 500 grams (1/4)^4/1
Nt = 2 grams
6. Nt = No (1/2)^x/k
= 100 grams (1/2)^168/12
= 100 grams (1/2)^14
Nt = 0.0061 grams
8. N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 years
9. N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
12. Pn = P(1+r)^n
= 75,000 (1+0.025)^ -15
P(-15)= 51,785 people
13. Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (1-0.10)^t
log (500,000/1,000,000) = t [log(1-0.10)]
t= log (500,000/1,000,000) / [log(1-0.10)]
t= 1.4 years
14. Pn = P(1+r)^n
500 = 100(1+0.05)^t
500/100 = (1+0.05)^t
log (500/100) = t [log(1+0.05)]
t= log (500/100) / [log(1+ 0.05)]
t = 32. 97 days
t= 33 days
Math Homework. p.355 09/13/12
ReplyDelete5.P= 8,000(1-.10/12)^12
Answer: P= P7,235.66
7.P= 10,000(1+.04/12)^24
Answer: P= P10,831.43
10.P= 100000(1+.09/4)^20
Answer: P= P156,050.92
Incomplete
Delete
ReplyDeleteAlexandra Cuartero
Written Math (p.355-356)
1. Pn = P(1+r)^n
= 50,000 (1+0.08)^4
P4 = 68,024 people
2. Nt = No (2)^x/k
= 100 bacteria (3)^6/1
Nt = 72,900 bacteria
3. Pn = 5000e^(0.15*8)
= 5000e^1.2 ≈ 5000(3.320116923 )
P8 = 16601 people
4. Nt = No (1/2)^x/k
= 500 grams (1/4)^4/1
Nt = 2 grams
6. Nt = No (1/2)^x/k
= 100 grams (1/2)^168/12
= 100 grams (1/2)^14
Nt = 0.0061 grams
8. N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 years
9. N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
12. Pn = P(1+r)^n
= 75,000 (1+0.025)^ -15
P(-15)= 51,785 people
13. Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (1-0.10)^t
log (500,000/1,000,000) = t [log(1-0.10)]
t= log (500,000/1,000,000) / [log(1-0.10)]
t= 1.4 years
14. Pn = P(1+r)^n
500 = 100(1+0.05)^t
500/100 = (1+0.05)^t
log (500/100) = t [log(1+0.05)]
t= log (500/100) / [log(1+ 0.05)]
t = 32. 97 days
t= 33 days
Math Homework p.356 numbers 5,7,10,13,15.
ReplyDeleteThe ones with ✓ are my final answers. :)
Solutions:
#5
A = P(1-r/n)^nt
= 8000(1-.10/12)(12)(1)
= ₱ 7,235.67 ✓
#7
A = P(1+r/n)^nt
= 10000(1+.04/12)^12(2)
= ₱ 10,831.43 ✓
#10
A = P(1+r/n)^nt
= 100,000(1+.09/4)^4(5)
= ₱ 10,831.43 ✓
#13
Q = p(1-d)^n
500,000 = 1,000,000(.9)^n (divide both sides by 1,000,000)
so, .5=(.9)^n
log.5/log.9
= 7 years ✓
#15
A = P(1+r/n)^nt
Banko Matatag
= 1,000,000(1+.06/2)^2(5)
= ₱ 1,343,916.38
Banko Subok
= 1,000,000(1+.05/4)^4(5)
= ₱ 1,282,037.23
I would deposit my money in Banko Matatag ✓
Assignment # 2 page 355 5,7,10,13,15
ReplyDeleteFristine Chua
Bk. Pg 356 5,7,10,13,15
5.) 8000(1-.10/12)(12)(1)
P= ₱ 7235.67
7.) 10,000 (1+.04/12)(2)(12)
P= ₱ 10,831.43
10.) 100,000(1+.09/4)(5)(4)
P= ₱ 156,050.92
13.) Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 years = 7 years
15.) Banko Matatag 1,000,000 (1+.06/2)(2)(5) = 1,343,916.379
Banko Subok 1,000,000 (1+.06/2)(2)(5) =1,280,084.544
Banko Matatag
Alexandra Cuartero 09/13/2012
ReplyDeleteAssignment (page 356, # 5, 7,10,13,15)
A= P (1+r/n)^nt
= 8000 (1-.10/12)^12(1)
= P 7,235.67
A= P (1+r/n)^nt
= 10,000 (1+.04/12)^12(2)
= P 10,831.43
10. A= P (1+r/n)^nt
= 100,000 (1+ .09/4)^4(5)
= P 156,050. 92
13. Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 years = 7 years
Banko Matatag:
A= P (1+r/n)^nt
= 1,000,000 (1+.06/2)^2(5)
= P 1,343,916.38
Banko Subok:
A= P (1+r/n)^nt
= 1,000,000 (1+.05/4)^4(5)
= P 1,282,037.23
I would deposit my money in Banko Matatag because I could earn as much as P 343,916. Unlike in Banko Subok where I could only earn P 282,037.
1. P=50,000 [1+.08]^12*4/6
ReplyDelete50,000 [ 1.08 ]^8
50,000 [1.85]
=92,500 people
2. N=100 [3]^6
=72,900 bacteria
3. P=5,000e^[.15*8]
P=5,000e^1.2
P=16,600 people
4. N=50[1/4]^4
=0.2 grams
6. N=100 [1/2]^14
=0.006 grams
8. 452=8[2]^t
log base 2 of 56.5=t
t=5.82 hrs
9. N=8[2]^10
=8 [1024]
=8,192 bacteria
12. 75,000 [1.025]^-15
51,784 people
13. 500,000=1,000,000 [1-0.1]^n
.5=.9^n
about 7 years
14. 500=100 [1+.05]^n
5=1.05^n
33 days
Edgar Caballero 10-Pascal
ReplyDeleteMath H.W
#1
(50000)(.08)=400
400*2=800
800*4=32000
50000+32000= 82000
pop = 82000
#2 Nt=No(2)^x/k
(100)(3)^6/1 = 72900
Nt = 72900 pesos
#4 Nt = No (1/2)^x/k
= (500g) (1/4)^4/1 = 2g
Nt = 2 grams
#5 At=Ao(1+r)^n
8000(1-10)^1 = 7200
7200 pesos
#6 Nt=No(1/2)^x/k
(100g)(1/2)^168/12
(100g)(1/2)^14= .0061g
NT = .0061g
#7 At=Ao(1+r)^n
10000(1+.04/12)^(12)(2) = 10831.43
10831.43 pesos
#8 N=(8)(2)^t
452 = 8*2^t
452/8 = 2^t = 56.5=2^t
log (56.5) = t *log(2)
t= log (56.5) / log(2) = 5.82
t= 5.82 or 6 years
#10 At=Ao(1+r)^n
100000(1+.09/4)^(5)(4) = 156050.92
156050.92 pesos
#13
50,000 = 1000000(1-0.10)^t
500000/1000000 = (1-0.10)^t
log (500000/1000000) = tlog(1-0.10)
t= log (500000/1000000) / log(1- 0.10) = 1.4
1.4 years
#15 At=Ao(1+r)^n
1000000(1+.06/2)^2 = 1060900 pesos
1000000(1+.05/4)^4 = 1050945.337 pesos
Banko Matatag
Math HW (p. 356)
ReplyDelete5) A = 8,000(1-.10/12)^12(1)
A = P 7,235.67
7) A = 10,000(1+.04/12)^12(2)
A = P 10,831.43
10) A = 100,000(1+.09/4)^4(5)
A = P 156,050.92
13) 500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t = log (1/2) / [log(.9)]
t = 6.58 years
t = 7 years
15) Banko Matatag:
A = 1,000,000(1+.06/2)^2(5)
A = P 1,343,916.38
Banko Subok:
A = 1,000,000(1+.05/4)^4(5)
A = P 1,282,037.23
I would deposit my money in Banko Matatag because they will
give me more money :)
Icomplete
DeleteMath HW for today page 356 #'s, 5,7,10,13,15
ReplyDelete(5) A= P (1+r/n)^nt
8,000(1-.10/12)^12(1)
= ₱ 7235.67
(7) A= P (1+r/n)^nt
10,000(1+.04/12)^12(2)
= ₱ 10,831.43
(10) A= P (1+r/n)^nt
100,000(1+.09/4)^4(5)
= ₱ 156,050.92
(13) Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 years
Final answer = 7 years
(15) A= P(1+r/n)^nt
Banko Matatag
1,000,000(1+.06/2)^2(5)
= ₱ 1,343,916.38
Banko Subok
1,000,000(1+.05/4)^4(5)
= ₱ 1,282,037.23
I would prefer to deposit my money in Banko Matatag
Sept. 13, 2012
ReplyDeletep. 356 #5, 7, 10, 13, 15
#5
A= P(1+r/n)^nt
=8000(1+[-10/12])^12(1)
= P 7235.67 ✓
#7
A= P(1+r/n)^nt
=10000(1+[.04/12])^12(2)
= P 10831.43 ✓
#10
A= P(1+r/n)^nt
=10000(1+[.09/4])^4(5)
= P 156050.92 ✓
#13
Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 years = 7 years ✓
#15
A= P(1+r/n)^nt
Banko Matatag = 1000000(1+[.06/2])^2(5)
= P 1343916.38 ✓
Banko Subok = 1000000(1+[.05/4])^4(5)
= P 1282037.23 ✓
I choose Banko Matatag ✓
95
DeleteHomework for today
ReplyDeleteProblem#5
Given:
P = 8000
r = 0.10
n = 12
t = 1
A = P(1+r/n)nt
= 8000(1-.10/12)^12
Answer= P 7235.67
Problem#7
Given:
P = 10000
r = 0.04
n = 12
t = 2
A = P(1+r/n)^nt
= 10000(1+0.04/12)24
Answer = P 10831.43
Problem#10
Given:
P = 100000
r = 0.09
n = 4
t = 5
A = P(1+r/n)^nt
= 100000(1+0.09/4)^20
Answer = P 156050.92
Problem#13
Given:
Q = 500000
P = 1000000
d = 0.10
Q = P(1-d)^n
500000 = 1000000(1-0.10)^n
500000 = 1000000(0.9)^n
* to find n, divide both sides by 1000000*
0.5 = 0.9^n
n= log (0.5)/log(0.9)
Answer = 6.47 = 7 years
Problem#15
Given:
P = 1000000
r1= 0.06
n1= 2
t = 5
r2= 0.05
n2= 4
A = P(1+r/n)^nt
Banko Matatag
= 1000000(1+0.06/2)^10
= 1000000(1.343916379)
Answer = 1343916.38
Banko Subok
= 1000000(1+0.05/4)^20
= 1000000(1.282037231)
Answer = 1282037.23
I would definitely deposit my money in Banko Matatag
MATH HOMEWORK as of SEPTEMBER 12 2012
ReplyDelete1.) Pn= P(1+r)^n
= 50,000(1+0.08)^4
P4= 68,024 people
There are 68,024 people after 4 years.
2.) Nt = No (2)^x/k
= 100 bacteria (3)^6
Nt= 72,900 bacteria
There will be 72,900 bacteria in the petri dish after 6 hours.
3.) Pn= 5000(2.718)^(0.15x8)
y= 2008-2000
y= 8
P8= 16601 people
The human population will be 16,601 at the year 2008
4.) Nt= No (1/4)^x/k
= 500g (1/4)^4
Nt= 2g
The substance weight would be 2g after 4 minutes.
5.) Nt= No (1/2)^x/k
= 100g(1/2)^14
Nt= 0.0061g
After one week, the remaining substance would be 0.0061g.
6.) N= 8(2)^t
452= 8(2)^t
452/8= (2)^t
Log(452/8)= t [log(2)]
T= 5.82 hours
The culture needs 5.82 hours to grow 452 bacteria.
7.) N= 8(2)^10
= 8,192 bacteria
There will be 8,192 bacteria after 10 hours.
8.) Q= 75000 (1+.025)^-15
= 75000 (.6904)
= 51,784.9 or 51,785
The human population of the province 15 years ago was 51,785.
9.) Q = P(1-d)^t
500000= 1000000(1-0.10)^t
500000/1000000= (1-0.10)^t
log (500000/1000000)= t [log(1-0.10)]
t= log (500,000/1,000,000) / [log(1-0.10)]
t= 1.4 years
It would take 1.4 years buy the car at half its price.
10.) Pn = P(1+r)n
500 = 100(1+0.05)^t
500/100 = (1+0.05)^t
log (500/100) = t [log(1+0.05)]
t= log (500/100)/[log(1+ 0.05)]
t = 32. 97 or 33 days
It would take the SARS virus 33 approximately 33 days to infect 500 people in China.
p.356
ReplyDeleteno.5
A=P(1+r/n)^nt
P=8000
r=.10
n=
t=1
8000(1+.10/12)^12
P=7235.67
no.7
A=P(1+r/n)^nt
P=10000
r=.04
n=12
t=2
10000(1+.04/12)^24
P=10831.43
no.10
A=P(1+r/n)^nt
P=100000
r=.09
n=4
t=5
100000(1+.09/4)^20
P= 156050.92
no.13
q=p(1-d)^n
P=1000000
d=.10
q=500000
500000=1000000(1-.10)^n
log1/2 + log .10 = n
n=2.3 yrs
no.15
A=P(1+r/n)^nt
P=1000000
r=.06
n=2
t=5
1000000(1+.06/2)^10
= 1343916.38
P=1000000
r=.05
n=4
t=5
1000000(1+.05/4)^20
= 1282037.23
Written Math (p.355-356)
ReplyDelete1. Pn = P(1+r)^n
= 50,000 (1+0.08)^4
P4 = 68,024 people
2. Nt = No (2)^x/k
= 100 bacteria (3)^6/1
Nt = 72,900 bacteria
3. Pn = 5000e^(0.15*8)
= 5000e^1.2 ≈ 5000(3.320116923 )
P8 = 16601 people
4. Nt = No (1/2)^x/k
= 500 grams (1/4)^4/1
Nt = 2 grams
6. Nt = No (1/2)^x/k
= 100 grams (1/2)^168/12
= 100 grams (1/2)^14
Nt = 0.0061 grams
8. N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 years
9. N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
12. Pn = P(1+r)^n
= 75,000 (1+0.025)^ -15
P(-15)= 51,785 people
13. Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (1-0.10)^t
log (500,000/1,000,000) = t [log(1-0.10)]
t= log (500,000/1,000,000) / [log(1-0.10)]
t= 1.4 years
14. Pn = P(1+r)^n
500 = 100(1+0.05)^t
500/100 = (1+0.05)^t
log (500/100) = t [log(1+0.05)]
t= log (500/100) / [log(1+ 0.05)]
t = 32. 97 days
t= 33 days
written math (page355)
5) A = 8,000(1-.10/12)^12(1)
A = P 7,235.67
7) A = 10,000(1+.04/12)^12(2)
A = P 10,831.43
10) A = 100,000(1+.09/4)^4(5)
A = P 156,050.92
13) 500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t = log (1/2) / [log(.9)]
t = 6.58 years
t = 7 years
15) Banko Matatag:
A = 1,000,000(1+.06/2)^2(5)
A = P 1,343,916.38
Banko Subok:
A = 1,000,000(1+.05/4)^4(5)
A = P 1,282,037.23
I would like to deposit my money to banko matatag because i can earn more haha
Math HW yesterday #'s 1,2,3,4,6,8,9,12 &14
ReplyDelete(1) P = 50,000 (1+ 0.08)^8/0.5
= 17,297.13 population of people
(2) Nt= 100(3)^6
=72,900 bacteria
(3) P= (5000)(2.71828)^(0.15)(8)
= 16,600.57 or 16,601 population of people
(4) Nt= 50(0.25)^4
= 0.1953125
Final answer= 0.20 grams
(6) Nt=100(1/2)^14
= 6.1035x10-3 or 6 bacteria
(8) 8 N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 hours
(9) N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
(12) Pt= 75000(1+0.025)^-15
= 51,784.91676 or 51,785 population of people
(13) Q= 1,000,000(1-0.10)^7
1,000,000(0.90)^7
478296.9
Final answer= 7 years
(14) 100(1+0.05)^33
500.32
Final answer= 33 days
1.) P= 8000(1 + -.10/12)^12(1)
ReplyDelete= 8000(0.904458)
= P7,235.66
2.) P= 10000(1 + .04/12)^12(2)
= P10,831.43
3.) P= 100000(1 + .09/4)^4(5)
= 1000000(1.5605)
P= 156,050.92
4.) 500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 or 7 years
5.) Banko Matatag:
P= 1000000(1 + .06/2)^2(5)
=1000000(1.3439)
P= P1,343,916.37
Banko Subok:
P= 1000000(1 + .05/4)^4(5)
= 1000000(1.2820)
P= P1,282,037.23
I would prefer to have my money in Banko Matatag.
5.)A=P(1-r/n)^nt
ReplyDelete=8000(1-.10/12)(12)(1)
ANSWER=P7,235.67
7.)A=P(1+r/n)^nt
=10000(1+.04/12)^12(2)
ANSWER=P10,831.43 ✓
10.)A=P(1+r/n)^nt
=100,000(1+.09/4)^4(5)
=P10,831.43
13.)Q=p(1-d)^n
500,000=1,000,000(.9)^n
.5=(.9)^n
log.5/log.9
ANSWER=7 years
15.)A=P(1+r/n)^nt
Banko Matatag
=1,000,000(1+.06/2)^2(5)
Banko Matatag=P1,343,916.38
Banko Subok
=1,000,000(1+.05/4)^4(5)
Banko Subok=P1,282,037.23
I will place my money in Banko Matatag.
Math HW yesterday #'s 1,2,3,4,6,8,9,12 &14
ReplyDelete(1) P = 50,000 (1+ 0.08)^8/0.5
= 17,297.13 population of people
(2) Nt= 100(3)^6
=72,900 bacteria
(3) P= (5000)(2.71828)^(0.15)(8)
= 16,600.57 or 16,601 population of people
(4) Nt= 50(0.25)^4
= 0.1953125
Final answer= 0.20 grams
(6) Nt=100(1/2)^14
= 6.1035x10-3 or 6 bacteria
(8) 8 N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 hours
(9) N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
(12) Pt= 75000(1+0.025)^-15
= 51,784.91676 or 51,785 population of people
(13) Q= 1,000,000(1-0.10)^7
1,000,000(0.90)^7
478296.9
Final answer= 7 years
(14) 100(1+0.05)^33
500.32
Final answer= 33 days
SEPTEMBER 13, 2012 HOMEWORK
ReplyDelete5.
A= P (1+r/n)^nt
8,000(1-.10/12)^12(1)
= ₱ 7235.67
7.
A= P (1+r/n)^nt
10,000(1+.04/12)^12(2)
= ₱ 10,831.43
10.
A= P (1+r/n)^nt
100,000(1+.09/4)^4(5)
= ₱ 156,050.92
13.
Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 years
Final answer = 7 years
15.
A= P(1+r/n)^nt
BM
1,000,000(1+.06/2)^2(5)
= ₱ 1,343,916.38
BK
1,000,000(1+.05/4)^4(5)
= ₱ 1,282,037.23
---> I choose Banko Matatag <---
HOMEWORK YESTERDAY, Sept. 12, 2012
ReplyDelete1. P= 50,000
r= 0.08
n= 4
initial time= 6 or 0.5
P4 = 50,000 (1+ 0.08) 8/0.5
17,297.13
2. 100 bacteria
3x every 60 mins
6 hours
P6 = 100(3)6
72,900
3. P= (5000)(2.71828)(0.15)(8)
16,600.57 or 16,601
4. P= 50(0.25)4
0.1953125 or 0
6. P=100(1/2)14
6.1035x10-3 or 6
8-9. 8(2)t
8 N = 8(2)^t
452 = 8(2)^t
452/8 = 2^t
log (452/8) = t [log(2)]
t= log (452/8) / log(2)
t= 5.82 hours
N = 8(2)^t
= 8(2)^10
N = 8,192 bacteria
12. 75000(1+0.025)-15
51,784.91676 or 51,785
13. 1,000,000(1-0.10)something
1,000,000(0.90)7
478296.9
7 years
14. 100(1+0.05)33
500.32 answer : 33 days
Math HW (p. 356)
ReplyDeleteCOMPLETE SOLUTION
5) A = P(1-r/n)^n(t)
A = 8,000(1-.10/12)^12(1)
A = P 7,235.67
7) A = P(1+r/n)^n(t)
A = 10,000(1+.04/12)^12(2)
A = P 10,831.43
10) A = P(1+r/n)^n(t)
A = 100,000(1+.09/4)^4(5)
A = P 156,050.92
13) Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t = log (1/2) / [log(.9)]
t = 6.58 years
t = 7 years
15) Banko Matatag:
A = P(1+r/n)^n(t)
A = 1,000,000(1+.06/2)^2(5)
A = P 1,343,916.38
Banko Subok:
A = P(1+r/n)^n(t)
A = 1,000,000(1+.05/4)^4(5)
A = P 1,282,037.23
I would deposit my money in Banko Matatag because they will give me more money.
Math hw p 355 5.7.10.13.15
ReplyDelete5)A=P(1+r/n)^nt
8,000(1-10/12)^(12)(1)
= p 7235.67
7)A=P(1+r/n)^nt
10.000(1-.04/12)^(12)(2)
= p 10,831,43
10) A=P(1+r/n)^nt
100,000 (1+.09/4)^(4)(5)
= P 156,050,92
13) Q=P(1-d)^t
500.000 =1,000,000 (1-0.10)^t
500.000 /1.000.000 =(. 9)^t
T=log(1/2)/[log(.9)]
T=6.58 years
Fa= 7years
15) A=P(1+r/n)^nt
Banco matatag
1.000.000 (1+.06/2)^(2)(5)
= P1,343,916,38
Banco subok
1,000,000 (1+.05/4)^(4)(5)
= P 1,282,037.23
My choice is banko matatag
09.13.12
ReplyDeletepage.356
(5) A= P (1+r/n)^nt
8,000(1-.10/12)^12(1)
= ₱ 7235.67
(7) A= P (1+r/n)^nt
10,000(1+.04/12)^12(2)
= ₱ 10,831.43
(10) A= P (1+r/n)^nt
100,000(1+.09/4)^4(5)
= ₱ 156,050.92
(13) Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t= log (1/2) / [log(.9)]
t= 6.58 years
Final answer = 7 years
(15) A= P(1+r/n)^nt
Banko Matatag
1,000,000(1+.06/2)^2(5)
= ₱ 1,343,916.38
5)A=P (1+r/n)^nt
ReplyDelete8,000 (1-.10/12)^(12)(1)
= p 7235.67
7) A =P (1+r/n)^nt
10.000 (1-.04/12)^(12)(2)
= p 10,831,43
10)A=P (1+r/n)^nt
100,000 (1+.9/4)^(12)(2)
= p 156,050,92
13) Q=P (1-d/n)^nt
500.000 =1.000.000
500.000 /1.000.000 =(.9)^nt
T=log(1/2)/[log(.9)]
T=6.58years
Fa = 7years
15) A=P (1+r/n)^n
Banko matatag
1.000.000 (1+.06/2)^(2)(5)
= P 1.343.916.38
Banko subok
1,000,000 (1+0.5/4)^(4)(5)
P = 1,282,037.23
MY CHOICE IS BANKO MATATAG
Homework for September 12, 2012.
ReplyDeletep. 355-356
#1
Pt = P(1+r)^n
= 50,000(1+0.16)^4
= 50,000(1.81)
Pt = 90,500 people
#2
Nt = No(2)^x/k
= 100(2)^6/1
= 100(64)
Nt = 6,400 bacteria
#3
Pt = 5,000(1+0.15)^8
= 5,000(3.06)
Pt = 15,300 people
#6
Nt = No(1/2)^x/k
= 100(1/2)^168/12
= 100(1/2)^14
= 100(1/16384)
Nt = 25/4096
#9
Nt = 452(2)^10/8
= 452(2.38)
Nt = 1076 bacteria
#12
Pt = 75,000(1+0.025)^15
= 75,000(1.45)
Pt = 108,750 people
#13
Pt = 5 years
A=p(1+r)^nt
ReplyDelete8,000 (1-.10/12)^(12)(1)
P = p7235.67
7) A =P (1+r/n)^nt
10.000 (1-0.4/12)^(12)(2)
= 108.622.36
10) A=P (1+r/n)^nt
100,000 (1+.09/4)^12(2)
= p 156,050,92
13) Q =P (1-d/n)^nt
500.000 = 1,000,000
500.000 /1,000,000 = (. 9)^t
T = log (1/2)/[log(.9)]
T = 6.58years
Fa = 7years
15) A =P (1+r/n) ^ n
Banko matatag
1.000.000 (1+.06/2)^(2)(5)
= P 1.343.916.38
Banko subok
1,000,000 ( 1+0.5/4)^,(4)(5)
= P 1,282,037.23
Homework for September 13, 2012.
ReplyDeletep. 356
#5
A = P(1+r/n)^nt
= 8000(1-0.1/12)^12(1)
= 8000(0.90)
A = 7,200 pesos
#7
A = 10,000(1+0.04/12)^12(2)
= 10,000(1)
A = 10,000 pesos
#10
A = 100,000(1+0.09/4)^4(5)
= 100,000(1.56)
A = 156,000 pesos
#15
P = 1,000,000(1+0.06/2)^2(5)
= 1,000,000(1.03)^10
= 1,000,000 (1.34)
P = 1,340,000 pesos
P = 1,000,000(1+0.05/4)^4(5)
= 1,000,000(1.01)^10
= 1,000,000(1.13)
P = 1,130,000 pesos
I would pick Banko Matatag because I would earn more money there unlike in Banko Subok.
5
ReplyDeleteA= 8000(1-0.1/12)^12(1)
= 8000(0.90)
A= 7,200 pesos
7
A= 10,000(1+0.04/12)^12(2)
= 10,000(1)
A= 10,000 pesos
10
A = 100,000(1+0.09/4)^4(5)
= 100,000(1.56)
A= 156,000
15
P = 1,000,000(1+0.06/2)^2(5)
= 1,000,000(1.03)^10
= 1,000,000 (1.34)
P= 1,340,000
P= 1,000,000(1+0.05/4)^4(5)
= 1,000,000(1.01)^10
= 1,000,000(1.13)
P= 1,130,000
HOMEWORK. LATE.
ReplyDelete5) A = P(1-r/n)^n(t)
A = 8,000(1-.10/12)^12(1)
A = P 7,235.67
7) A = P(1+r/n)^n(t)
A = 10,000(1+.04/12)^12(2)
A = P 10,831.43
10) A = P(1+r/n)^n(t)
A = 100,000(1+.09/4)^4(5)
A = P 156,050.92
13) Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t = log (1/2) / [log(.9)]
t = 6.58 years
t = 7 years
15) Banko Matatag:
A = P(1+r/n)^n(t)
A = 1,000,000(1+.06/2)^2(5)
A = P 1,343,916.38
Banko Subok:
A = P(1+r/n)^n(t)
A = 1,000,000(1+.05/4)^4(5)
A = P 1,282,037.23
5.) P(1+r/n)^nt
ReplyDelete= 8000(1-0.1/12)^12(1)
= 8000(0.90)
= P7,200
7.) 10,000(1+0.04/12)^12(2)
= 10,000(1)
= P10,000
10.) 100,000(1+0.09/4)^4(5)
= 100,000(1.56)
= P156,000
15.) 1,000,000(1+0.06/2)^2(5)
= 1,000,000(1.03)^10
= 1,000,000 (1.34)
= P1,340,000
1,000,000(1+0.05/4)^4(5)
= 1,000,000(1.01)^10
= 1,000,000(1.13)
= P1,130,000
LATE HOMEWORK
ReplyDelete5) A = P(1-r/n)^n(t)
A = 8,000(1-.10/12)^12(1)
A = P 7,235.67
7) A = P(1+r/n)^n(t)
A = 10,000(1+.04/12)^12(2)
A = P 10,831.43
10) A = P(1+r/n)^n(t)
A = 100,000(1+.09/4)^4(5)
A = P 156,050.92
13) Q = P(1-d)^t
500,000 = 1,000,000(1-0.10)^t
500,000/1,000,000 = (.9)^t
t = log (1/2) / [log(.9)]
t = 6.58 years
t = 7 years
15) Banko Matatag:
A = P(1+r/n)^n(t)
A = 1,000,000(1+.06/2)^2(5)
A = P 1,343,916.38
Banko Subok:
A = P(1+r/n)^n(t)
A = 1,000,000(1+.05/4)^4(5)
A = P 1,282,037.23
KLAUDINE S. ALEJAR
ReplyDeleteOCTOBER 1, 2012
MATH AND I
Throughout the high school years, every student has their own opinions about the subjects they like and dislike. This does not mean the literal meaning. It means that it is about their strengths and weaknesses when it comes to academics. These opinions often change throughout the time. There are certain standards in every individual in order to succeed this kind of high school life.
“Mathematics is an essential part of every high school student’s education.” This quote says it all. It explains the subject Math in eleven words. Every student learns certain lessons in math that will carry with them through life. Each of these are building blocks to help shape who we become. As a math student, I feel math has taught me not only about addition, subtraction, division... But it has taught me thinking skills and problem solving that will take me further than just math class. There are times in which you have to decide which path you will take after high school, although every path is different and you will take the skills in which you have learned in math class to a different level. When I was young, I used to say “I hate math”, although this sentence did not last long. As I was growing up, I slowly understood what math really meant and how we can apply it in real life situations. I used to say that this has nothing to do with shopping or dancing and such and that I won’t need this in the future. For my other batchmates, these are hard, most especially when you are surrounded by your best friends. Concentration is obviously needed when it comes to this matter. The dedication that this class requires teaches every student concentration. The concentration on this one course will help high school students pursue their dreams while taking other courses to achieve this. Another major influence on a high school student’s math education is the teacher. A teacher is crucial to a student’s success. The ideal teacher knows several ways to explain a certain concept. Doc Ed is an example of an ideal teacher.
Soon enough, I know that I will become successful because of math. I know that this will be a big help in my future. I just have to focus at this moment and I can’t be distracted by the things that will destroy my future.