Answer:
2.60
Step-by-step explanation:
Given that:
n = 1600


The test statistics can be computed as:





 
        
             
        
        
        
Answer:
To completely fill the sandbox will cost $56
Step-by-step explanation:
First, we calculate volume of the rectangular-shaped sandbox:

We know that density of sand is 100 pounds per cubic foot. Then, to fill the sand box that has a volume of 8 cubic foot, we calculate by rule 3:
1cubic foot..............100 pounds
8 cubic foot.............x pounds


We will need 800 pounds of sand to fill the sandbox, then we can calculate by rule 3 the number of bags needed and next the total cost of them:
50 pounds.......1bag
800 pounds.....x bags

To calculate the cost:
1 bag ..................$3.50
16 bags.................$x

To completely fill the sandbox will cost $56
 
        
             
        
        
        
Answer:

Step-by-step explanation:
So we have the expression: 

And we wish to factor it. 
First, let's make a substitution. Let's let u be equal to x². Therefore, our expression is now: 

This is a technique called quadratic u-substitution. Now, we can factor in this form. 
We can use the numbers -3 and -2. So: 

For the first two terms, factor out a u. 
For the last two terms, factor out a -3. So: 

Grouping: 

Now, substitute back the x² for u: 

And this is the simplest form. 
And we're done!
 
        
             
        
        
        
9514 1404 393
Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
   f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
   f(0) = 2
   f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
   f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.