<h3>
Answer:</h3>
$450
<h3>
Step-by-step explanation:</h3>
Each year, the amount of interest earned is ...
... 4.5% × $5000 = $225
After 2 years, a total of ...
... $225 × 2 = $450
has been transferred.
_____
<em>Comment on percentages</em>
Of course, you know that 4.5% = 4.5/100 = 45/1000 = 0.045.
You can think of the % symbol as a fancy (shorthand) way to write /100. (Likewise, the ‰ symbol means /1000.)
Answer:
Let X = score of final exam.
X~normal(23, 16)
(a)
percentage of score between 19 and 27
= P(19 < X < 27)
= P((19-23) / sqrt(16) < Z < (27 - 23) / sqrt(16))
= P(-1 < Z < 1)
= 1 - P(Z <= -1) - P(Z >= 1)
= 1 - P(Z >= 1) - P(Z >= 1)
= 1 - 2*P(Z >= 1)
= 1 - 2(0.1587)
= 0.6826
= 68.26%
According to Normal Distribution Table
P(Z>=1) = 1 - P(Z<1) = 1-0.8413
So the final percentage is 68.26%
Answer: The required solution of the given differential equation is
Step-by-step explanation: We are given to solve the following differential equation :
Let, 
be an auxiliary solution of equation (i).
Then, 
Substituting these values in equation (i), we get
![m^3e^{mx}+4m^2e^{mx}-16me^{mx}-64e^{mx}=0\\\\\Rightarrow (m^3+4m^2-16m-64)e^{mx}=0\\\\\Rightarrow m^3+4m^2-16m-64=0,~~~~~~~~~[\textup{since }e^{mx}\neq 0]\\\\\Rightarrow m^2(m-4)+8m(m-4)+16(m-4)=0\\\\\Rightarrow (m-4)(m^2+8m+16)=0\\\\\Rightarrow (m-4)(m+4)^2=0\\\\\Rightarrow m-4=0,~~(m+4)^2=0\\\\\Rightarrow m=4,~m=-4,~-4.](https://tex.z-dn.net/?f=m%5E3e%5E%7Bmx%7D%2B4m%5E2e%5E%7Bmx%7D-16me%5E%7Bmx%7D-64e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E3%2B4m%5E2-16m-64%29e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E3%2B4m%5E2-16m-64%3D0%2C~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmx%7D%5Cneq%200%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%28m-4%29%2B8m%28m-4%29%2B16%28m-4%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m-4%29%28m%5E2%2B8m%2B16%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m-4%29%28m%2B4%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m-4%3D0%2C~~%28m%2B4%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D4%2C~m%3D-4%2C~-4.)
So, the general solution is given by
Then, we have
With the conditions given, we get
![y^\prime(0)=4A-4B+C\\\\\Rightarrow 4A-4B+C=26\\\\\Rightarrow 4(A+A)+C=26~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow C=26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D4A-4B%2BC%5C%5C%5C%5C%5CRightarrow%204A-4B%2BC%3D26%5C%5C%5C%5C%5CRightarrow%204%28A%2BA%29%2BC%3D26~~~~~~~~~~~~~~~~%5B%5Ctextup%7Busing%20equation%20%28i%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20C%3D26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~%28iii%29)
and
![y^{\prime\prime}(0)=16A+16B-8C\\\\\Rightarrow 16A-16A-8C=-16~~~~~~~~~~~~[\textup{using equation (ii)}]\\\\\Rightarrow -8C=-16\\\\\Rightarrow C=2.](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%5Cprime%7D%280%29%3D16A%2B16B-8C%5C%5C%5C%5C%5CRightarrow%2016A-16A-8C%3D-16~~~~~~~~~~~~%5B%5Ctextup%7Busing%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-8C%3D-16%5C%5C%5C%5C%5CRightarrow%20C%3D2.)
From equation (iii), we get
From equation (ii), we get
Therefore, the required solution of the given differential equation is
That would cost her $25.02
Unsure what the T/F statements are they are not included
The second one is your answer
