The minimum surface area that such a box can have is 380 square
<h3>How to determine the minimum surface area such a box can have?</h3>
Represent the base length with x and the bwith h.
So, the volume is
V = x^2h
This gives
x^2h = 500
Make h the subject
h = 500/x^2
The surface area is
S = 2(x^2 + 2xh)
Expand
S = 2x^2 + 4xh
Substitute h = 500/x^2
S = 2x^2 + 4x * 500/x^2
Evaluate
S = 2x^2 + 2000/x
Differentiate
S' = 4x - 2000/x^2
Set the equation to 0
4x - 2000/x^2 = 0
Multiply through by x^2
4x^3 - 2000 = 0
This gives
4x^3= 2000
Divide by 4
x^3 = 500
Take the cube root
x = 7.94
Substitute x = 7.94 in S = 2x^2 + 2000/x
S = 2 * 7.94^2 + 2000/7.94
Evaluate
S = 380
Hence, the minimum surface area that such a box can have is 380 square
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Answer:
Step-by-step explanation:
n = 400
Proportion p = 229/400= 0.5725
For 95% confidence interval we use Z value as 1.96
Std error = 0.025
margin of error = 1.96*0.025
Confidence interval 95% = 0.5725±Margin of error
= (0.524, 0.621)
b) When smiled x becomes 277
p = 0.6925
Std error = 0.023
Margin of error= 1.96*0.023
Confidence interval = (0.647, 0.738)
Smiling increases the chances of stopping since mean and conidence interval bounds are showing increasing trend.
1. 7 17/42
2. 42 5/6
3. 7 3/4
4. 12 2/3
5. 2 8/35
The answer is 11abc^2 + 12a^2b + 18b^2c because:
14abc^2-3abc^2 is 11abc^2
12a^2b + 0 = 12a^2b
16b^2c +2b^2c = 18b^c
First, subtract y2 - y1 to find the vertical distance. Then, subtract x2 - x1 to find the horizontal distance.
Formula to find distance given two points.
Square root (X2 - X1)^2 + (Y2 - Y1)^2
Xa Ya Xb Yb
A = (3, -4) B = (-1, 3)
Xa goes into X2 and Xb goes into X1
(3 - (-1))^2
Ya goes into Y2 and Yb goes into Y1
(-4 - 3)^2
Square root (3 - (-1))^2 + (-4 - 3)^2
Square root (4)^2 + (-7)^2
Square root 16 + 49
Square root 65
= 8.06
The error was Drako had (3 - 4)^2 when it should have been (3 - (-4))^2 because a positive is subtracting a negative.
