Answer:
<u>32 53 24 54 61 13</u>
<u>32 53 24 54 61 1317 36 04 10 51 68</u>
Answer:
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Step-by-step explanation:
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Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
Answer:
Since the calculated value of F = 1.4397 is less than the critical value of
F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.
Step-by-step explanation:
1)Formulate the hypothesis that first variance is equal or greater than the second variance
H0: σ₁²≥σ₂² against the claim that the first instructor's variance is smaller
Ha: σ₁²< σ₂²
2) Test Statistic F= s₂²/s₁²
F= 84.8/ 58.9= 1.4397
3)Degrees of Freedom = n1-1= 10-1= 9 and n2 = 10-1= 9
4)Critical value at 10 % significance level= F(9,9)= 2.4403
5)Since the calculated value of F = 1.4397 is less than the critical value of
F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.
September 29, 2016 is today's date.
