<span>E[Y] = 0.4·1 + 0.3·2 + 0.2·3 + 0.1·4 = 2
E[1/Y] =0.4·1/1 + 0.3·1/2 + 0.2·1/3 + 0.1·1/4 = 0.4 + 0.15 + 0.0666 + 0.025?0.64
V[Y] =E[Y2]-E[Y]2= (0.4)·12+(0.3)·22+(0.2)·32+(0.1)·42-22= 0.4+1.2+1.8+1.6-4= 5-4 = 1</span>
First, it would help to simplify each side more:
Left side: 36 + 3(4x - 9) = 36 + 12x - 27 = 12x + 9
Right side: c(2x + 1) + 25 = 2cx + c + 25
Write the simplified equation:
12x + 9 = 2cx + c + 25
Usually when there is no solution, the coefficients of the variable on both sides are the same, so we can make the coefficient if x on the right side into 12:
2cx >>> 12x
Then, c must equal 6 to make this true.
The answer is choice (C).
Answer:
<h2><em>
2ft by 2ft by 1 ft</em></h2>
Step-by-step explanation:
Total surface of the cardboard box is expressed as S = 2LW + 2WH + 2LH where L is the length of the box, W is the width and H is the height of the box. Since the cardboard box is without a lid, then the total surface area will be expressed as;
S = lw+2wh+2lh ... 1
Given the volume V = lwh = 4ft³ ... 2
From equation 2;
h = 4/lw
Substituting into r[equation 1;
S = lw + 2w(4/lw)+ 2l(4/lw)
S = lw+8/l+8/w
Differentiating the resulting equation with respect to w and l will give;
dS/dw = l + (-8w⁻²)
dS/dw = l - 8/w²
Similarly,
dS/dl = w + (-8l⁻²)
dS/dw = w - 8/l²
At turning point, ds/dw = 0 and ds/dl = 0
l - 8/w² = 0 and w - 8/l² = 0
l = 8/w² and w =8/l²
l = 8/(8/l² )²
l = 8/(64/I⁴)
l = 8*l⁴/64
l = l⁴/8
8l = l⁴
l³ = 8
l = ∛8
l = 2
Hence the length of the box is 2 feet
Substituting l = 2 into the function l = 8/w² to get the eidth w
2 = 8/w²
1 = 4/w²
w² = 4
w = 2 ft
width of the cardboard is 2 ft
Since Volume = lwh
4 = 2(2)h
4 = 4h
h = 1 ft
Height of the cardboard is 1 ft
<em>The dimensions of the box that requires the least amount of cardboard is 2ft by 2ft by 1 ft</em>
Answer:
-11/8
Step-by-step explanation:
-7/8 - 1/2 = -11/8
