What is the perimeter of the are. Please show work.
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Answer:
The rate at which water is being pumped into the tank is 13,913.27 cubic centimetre per minute
Step-by-step explanation:
Given that, the height of the conical tank is 13 meters and diameter is 3 meters.
Radius = meters
......(1)
The volume of the cone is
Now putting
Differentiating with respect to t
Given that, the water level is rising at a rate of 17 cm per minute when the height 1 meter= 100 cm .i.e cm/min
Putting
≈7,113 cubic centimetre per minute
The volume of water increases 7,113 cubic centimetre per minute while 6,800 cubic centimetre per minute is leaking out.
It means the required rate at which water is being pumped into the tank is
=(7,113+6,800)cubic centimetre per minute
=13,913 cubic centimetre per minute
(a) For any probability distribution, the total probability must be 1. That is, the area under the probability density curve must be equal to 1.
The empirical rule for normal distributions says
• approximately 68% of the distribution lies within 1 standard deviation of the mean
• approx. 95% lies within 2 s.d. of the mean
• approx. 99.7% lies within 3 s.d. of the mean
In this case, with mean 3500 and s.d. 470, this translates to
• Pr(3500 - 470 < X < 3500 + 470) = Pr(3030 < X < 3970) ≈ 0.68
• Pr(3500 - 2*470 < X < 3500 + 2*470) = Pr(2560 < X < 4440) ≈ 0.95
• Pr(3500 - 3*470 < X < 3500 + 3*470) = Pr(2090 < X < 4910) ≈ 0.997
Continuous probability distributions also have the property that
Pr(a < X < b) = Pr(a < X < c) + Pr(c < X < b)
if a < c < b.
Combining all these properties, we can find the probabilities for each of the 8 regions in the graph to be (from left to right)
• Pr(-∞ < X < 2090) ≈ (1 - 0.997)/2 ≈ 0.0015
• Pr(2090 < X < 2560) ≈ (1 - 0.95 - 2*0.0015)/2 ≈ 0.0235
• Pr(2560 < X < 3030) ≈ (1 - 0.68 - 2*0.0235 - 2*0.0015)/2 ≈ 0.135
• Pr(3030 < X < 3500) ≈ 0.68/2 ≈ 0.34
and since the distribution is symmetric about its mean, we already know the remaining probabilities,
• Pr(3500 < X < 3970) ≈ 0.34
• Pr(3970 < X < 4440) ≈ 0.135
• Pr(4440 < X< 4910) ≈ 0.0235
• Pr(4910 < X < ∞) ≈ 0.0015
(b) Per the rule, 99.7% of babies would weight between 2090 and 4910 grams.
(c) The proportion of babies weighing less than 3030 grams is the sum of the proportions of babies weighing less than 2090, between 2090 and 2560, and between 2560 and 3030 grams. So
Pr(X < 3030) = Pr(-∞ < X < 2090) + Pr(2090 < X < 2560) + P(2560 < X < 3030)
Pr(X < 3030) ≈ 0.0015 + 0.0235 + 0.135
Pr(X < 3030) ≈ 0.16 = 16%
(d) Similarly,
Pr(X > 2560) = Pr(2560 < X < 3030) + Pr(3030 < X < 3500) + … + Pr(4910 < X < ∞)
Pr(X > 2560) ≈ 0.135 + 0.34 + 0.34 + 0.135 + 0.0235 + 0.0015
Pr(X > 2560) ≈ 0.975 = 97.5%
