So here we use the pythagorean theorem which is a2 + b2 = c2 (“a” squared times “b” squared equals “c” squared) the length from calvins house to the intersection is “a” and the length from phoebes house to the intersection is “b” so in order to find out the length of “c” (calvins house to phoebes house) we need to use the pythagorean theorem 
a2 275x275=75,625
b2 113x113 = 12,769
so now that we have figured out what a2 and b2 are let’s add them
75,625 + 12,769 = 88,394 
now since we only need to find out the length and not the area we need to find the square root of 88,394 
the square root of 88,394 is 297.311 (i cut off the decimal after three places) 
the direct length from phoebes house to calvins is 297.311 meters
        
             
        
        
        
That's D.
A(t) is the amount as a function of the temperature t.
 
        
                    
             
        
        
        
According to the problem, the length is 2 feet more than its width, which is expressed as the following function

If the width is x, then the length is x+2.
We can express the following function for the area

Let's replace the width x = 11 ft.

<h2>Hence, the area is 143 square feet. </h2><h2>The function for the area is</h2>

Where x represents the width.
 
        
             
        
        
        
Answer:
p value = 0.302
Step-by-step explanation:
Given that all cars can be classified into one of four groups: the subcompact, the compact, the midsize, and the full-size. There are five cars in each group. Head injury data (in hic) for the dummies in the driver's seat are listed below. 
H_0: All cars have same mean values
H_a: atleast two cars have different mean values
(Two tailed anova test)
Anova: Single Factor      
      
SUMMARY      
Groups	Count	Sum	Average	Variance  
Subcompact	5	3444	688.8	48502.7  
compact	5	2879	575.8	4582.7  
Midsize	5	2534	506.8	18720.2  
Full size	5	2689	537.8	23905.2  
      
      
ANOVA      
Source of Variation	SS	df	MS            F             P-value	F crit
Between Groups	94825	3	31608.33	1.321    0.302           3.24
Within Groups	382843.2	16	23927.7    
      
Total	477668.2	19    
Since p value of 0.302 is greater than 0.05 significance level we accept null hypothesis.
p value = 0.302