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

Answer:
The degrees of freedom, Df = The number of bags produced on Monday - 1
Step-by-step explanation:
The number of degrees of freedom is the limiting number of values that are logically not influenced by other values such that they are capable of having variation
The degrees of freedom = The sample size - 1 = N - 1
Therefore, the degrees of freedom, Df = The number of bags produced on Monday - 1
Answer: 160 calories
Step-by-step explanation:
Answer:
x = 1
Step-by-step explanation:
Answer:
Mean = 8
Variance = 7.36
Standard Deviation = 2.7129
Step-by-step explanation:
This is a binomial distribution with parameters, n and p.
Where
n is sample size (given as 100)
p is the probability of success, or probability of defective (given as 8% or 0.08)
The mean, variance, standard deviation formula for binomial distribution is shown below:
Mean = 
Variance = 
Standard Deviation = 
Where q would be probability of failure, or "1 - p"
Thus,
n = 100
p = 0.08
q = 1 - 0.08 = 0.92
SO, we have:
Mean = 
Variance = 
Standard Deviation = 