The area of the region that is not shaded in the diagram is determined through the following solution:
<span>A = <span>(1 − (<span><span>60°/</span><span>360°</span></span>)*</span>π) * (10 ft ) ^<span>2</span></span>
<span>A = <span>56*</span>π*100 [<span>ft2</span>]</span>
<span><span>A = <span>250/3 </span>π [<span>ft2</span>]
A = 262 [ft2]</span></span>
Can you reformat the question?
The digit in the ten millions place of a number that is less than 55,000,000 but greater than 25,000,000 can be 3 or 4.
25,000,000 < <u>3</u>5,000,000 < 55,000,000
25,000,000 < <u>4</u>5,000,000 < 55,000,000
The sample size must be 543 so the sample proportion is will not differ by 5%
We are given the following in the question:
Margin of error = 5%
Confidence interval:
p ±z√(p'(1 - p')/n)
Margin of error =
p = ±z√(p'(1 - p')/n)
Since no particular proportion is given, we take
p' = 0.5
Z critical at α 0.02 is ± 2.33
Putting values, we get,
p = ±z√(p'(1 - p')/n)
2.33 x √(0.5(1 - 0.5)/n)
√n = 2.33x 0.5/0.05
n = 542.89 = 543
Thus, the sample size must be 543 so the sample proportion is will not differ by 5%
To learn more about confidence interval refer here
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Let L be the low copper alloy and H be the high copper alloy.
We need 0.15L + 0.60H = 0.42 and L + H = 100
L=100 - H
Substituting this into the second equation for L, we get:
0.15(100-H) + 0.60H = 0.42(H+L)
15 - 0.15H + 0.60H = 0.42(H+100-H)
15 + 0.45H = 0.42(100)
0.45H = 42 - 15 = 27
H = 27/.45 = 60 lbs
Back substitution, L = 40 lbs.
