Answer:
The correct answer is At least two of the stimulants will have different effects on the mean time spent awake.
Explanation:
The null hypothesis exposed leads to consider an alternative hypothesis that differs from the behavior of the 4 stimulants. In this case it is related to the effects on patients, since in practice at least one will present differences compared to the others, influencing the sleep time of consumers.
 
        
             
        
        
        
Answer:
Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.
 
        
                    
             
        
        
        
Answer:
a) at T = 5800 k  
   band emission = 0.2261
at T = 2900 k
   band emission = 0.0442
b) daylight (d) = 0.50 μm
     Incandescent ( i ) =  1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as 
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T ) 
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k  
   band emission = 0.2261
at T = 2900 k
   band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u> 
For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
 
        
             
        
        
        
Answer:
It would take approximately 305 s to go to 99% completion
Explanation:
Given that:
y = 50% = 0.5
n = 1.7
t = 100 s 
We need to first find the parameter k from the equation below.

taking the natural logarithm of both sides:

Substituting values:

Also
 ![t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}](https://tex.z-dn.net/?f=t%5En%3D-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%5C%5Ct%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D)
Substituting values and y = 99% = 0.99
![t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s](https://tex.z-dn.net/?f=t%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D%3D%5Csqrt%5B1.7%5D%7B-%5Cfrac%7Bln%281-0.99%29%7D%7B2.76%2A10%5E%7B-4%7D%7D%7D%3D304.6s)
∴ t ≅ 305 s
It would take approximately 305 s to go to 99% completion