For this question you would have to expand the numbers to the thounsandths
so it can be 9.181, 9.182, 9.183 and so on
Answer:
x = -0.3
Step-by-step explanation:
You can substitute y in 2x – 4y = 3, to solve for x.
How to solve for x:
2x - 4(3x) = 3 (Simplify)
2x - 12x = 3 (Simplify)
-10x = 3 (Divide by -10)
x = -0.3
Answer:
1
The probability is 
2
The probability is 
Step-by-step explanation:
From the question we are told that
The population mean is 
The standard deviation is 
The sample size is 
Generally the standard error for the sample mean
is mathematically evaluated as

substituting values


Apply central limit theorem[CLT] we have that
![P(\= X < 33) = [z < \frac{33 - \mu }{\sigma_{\= x}} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3C%20%20%5Cfrac%7B33%20-%20%20%5Cmu%20%7D%7B%5Csigma_%7B%5C%3D%20x%7D%7D%20%5D)
substituting values
![P(\= X < 33) = [z < \frac{33 - 28.29 }{4.48} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3C%20%20%5Cfrac%7B33%20-%20%2028.29%20%7D%7B4.48%7D%20%5D)
![P(\= X < 33) = [z < 1.05 ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3C%20%201.05%20%5D)
From the z-table we have that

For the second question
Apply central limit theorem[CLT] we have that
![P(\= X > 30 ) = [z > \frac{30 - \mu }{\sigma_{\= x}} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3E%2030%20%29%20%3D%20%20%5Bz%20%3E%20%20%5Cfrac%7B30%20-%20%20%5Cmu%20%7D%7B%5Csigma_%7B%5C%3D%20x%7D%7D%20%5D)
substituting values
![P(\= X < 33) = [z > \frac{30 - 28.29 }{4.48} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3E%20%20%5Cfrac%7B30%20-%20%2028.29%20%7D%7B4.48%7D%20%5D)
From the z-table we have that

Thus


Answer: Similar - AA
Just took the quiz, Its Similar AA not sure why just trust me homie