Answer:
the sample should be 237
Step-by-step explanation:
The computation of the large the sample be used is shown below:
Given that
Standard deviation = 0.09 ounces
margin of error = 0.011
For 94% confidence interval, the z value be 1.88
As we know that
Margin of error = ((z score × standard deviation) ÷ (Sample size))^2
0.011 = (1.88 × 0.09) ÷ (Sample size))^2
0.011 = 0.1692 ÷ (Sample size))^2
So, the sample size be
= 15.38^2
= 236.60
Hence, the sample should be 237
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:
I think 20 but i may of done my math wrong
Step-by-step explanation:
#5)
14/4 = 3.5
24.5 / 7 = 3.5
The pair is similar and SF = 3.5
#6)
10/4 = 2.5
21/9 = 2.3333
The pair is not similar because the corresponding sides are not proportional.
a. 1.15
b. 6.20 divided by 4 = 1.55
c. 5 divided by 4 = 1.25
d. 5.50 divided by 5 = 1.1
D is correct