#1. 712.5
#2. 1185
#3 .517.5
Answer:
r = 4.5%
Step-by-step explanation:
Formula:
I = Prt
r = I/(Pt)
Given:
I = 662.29
P = 4205
t = 3.5
Work:
r = I/(Pt)
r = 662.29/(4205 * 3.5)
r = 662.29/14717.5
r = 0.045
r = 4.5%
Answer:
2201.8348 ; 3 ; x / (1 + 0.01)
Step-by-step explanation:
1)
Final amount (A) = 2400 ; rate (r) = 6% = 0.06, time, t = 1.5 years
Sum = principal = p
Using the relation :
A = p(1 + rt)
2400 = p(1 + 0.06(1.5))
2400 = p(1 + 0.09)
2400 = p(1.09)
p = 2400 / 1.09
p = 2201.8348
2.)
12000 amount to 15600 at 10% simple interest
A = p(1 + rt)
15600 = 12000(1 + 0.1t)
15600 = 12000 + 1200t
15600 - 12000 = 1200t
3600 = 1200t
t = 3600 / 1200
t = 3 years
3.)
A = p(1 + rt)
x = p(1 + x/100 * 1/x)
x = p(1 + x /100x)
x = p(1 + 1 / 100)
x = p(1 + 0.01)
x = p(1.01)
x / 1.01 = p
x / (1 + 0.01)
2 years will be 80 cats (20 x 2 = 40 (year 1) and 40 x 2 = 80 (year 2))
3 years will be 160 (80 x 2 = 160)
5 years will be (160 x 2 = 320 (year 4) and 320 x 2 = 640 (year 5)) So 5 years will be 640
Answer:(a) margin error = 2.4%
(b) The margin error gives the measure in percentage of how the population parameter determined differ from the real population statistics or value.
(c) in 90% of the samples of teens in the country, the percent who go online several times a day will be within 50.6% and 55.4%. of the estimated 100%
Step-by-step explanation:
Using the proportion formulae
Margin error = z √p(1-p)/n
n= 1170, p = 53% = 0.53, 1-p = 0.47
and the z value at 90% C.I = 1.645
M error= 1.645 √0.53×0.47/1170
Margin error = 0.024 = 0.024 ×100
Margin error = 2.4%
53 - 2.4 = 50.6% and 53 + 2.4 = 55.4%
In other words 90% of the time: the number of teens who go online several time a day will be between 50.6 and 55.4%.
