If u tell me the solution's like are u evaluating or what, which method is it i can tell u the answer  <span />
        
             
        
        
        
Answer:
The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean  and standard deviation
 and standard deviation  , the zscore of a measure X is given by:
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the reading speed of a sixth-grader whose reading speed is at the 90th percentile
This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.




The reading speed of a sixth-grader whose reading speed is at the 90th percentile is 155.72 words per minute.
 
        
                    
             
        
        
        
Answer with explanation:

--------------------------------------------------------Dividing both sides by 8 x
This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.
Integrating Factor
   
 
Multiplying both sides by Integrating Factor  
 ![x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5Ctimes%20%5By%27%2By%5Ctimes%5Cfrac%7B1%2B4x%5E2%7D%7B8x%7D%5D%3D%5Cfrac%7B1%7D%7B8%7D%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5C%5C%5C%5C%20%5Ctext%7BIntegrating%20both%20sides%7D%5C%5C%5C%5Cy%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D-%5Bx%5E%7B%5Cfrac%7B9%7D%7B8%7D%7D%5D%5Ctimes%5Cfrac%7B%20%5CGamma%280.5625%2C%20-x%5E2%29%7D%7B%28-x%5E2%29%5E%7B%5Cfrac%7B9%7D%7B16%7D%7D%7D%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%28-1%29%5E%7B%5Cfrac%7B-1%7D%7B8%7D%7D%5B%20%5CGamma%280.5625%2C%20-x%5E2%29%5D%2BC-----%281%29)
When , x=1, gives , y=9.
Evaluate the value of C and substitute in the equation 1.
 
        
             
        
        
        
Plug in -3 for x
f(-3) = 2(-3) - (-3 + 6)
f(-3) = -6 - (3)
f(-3) = -9
Solution: f(-3) = -9