F=ir^t
139=134r^10
139/134=r^10
r=(139/134)^(1/10) then:
f=134(139/134)^(t/10) so in 2014, t=24 so
f=134(139/134)^(2.4)
f≈146 million (to nearest million)
Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...
A=Pe^(kt)
139=134e^(10k)
139/134=e^(10k)
ln(139/134)=10k
k=ln(139/134)/10 so
A=134e^(t*ln(139/134)/10) when t=24
A=134e^(2.4*ln(139/134))
A≈146 million (to nearest million)
The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
f(3) = 1
substitute x = 3 into f(x)
f(3) = (2 × 3) - 5 = 6 - 5 = 1
Answer: (0.076, 0.140)
Step-by-step explanation:
Confidence interval for population proportion (p) is given by :-
, where 
= sample proportion.
n= sample size.
= significance level .
= critical z-value (Two tailed)
As per given , we have
sample size : n= 500
The number of Independents.: x= 54
Sample proportion of Independents
Significance level 98% confidence level : 
By using z-table , Critical value : 
The 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters will be :-
Hence, the 98% confidence interval for the true percentage of Independents among Haywards 50,000 registered voters.= (0.076, 0.140)
E=9h+1.5(9)(h-40)
e=9h+13.5(h-40), since h=49
e=9(40)+13.5(9)
e=9(40+13.5)
e=9(53.5)
e=$481.50
