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lesya [120]
3 years ago
13

Boat A leaves a dock headed due east at 2:00PM traveling at a speed of 9 mi/hr. At the same time, Boat B leaves the same dock tr

aveling due south at a speed of 15 mi/hr. Find an equation that represents the distance d in miles between the boats and any time t in hours.
Mathematics
1 answer:
netineya [11]3 years ago
7 0

Answer: 17.5t

Step-by-step explanation:

Given

Speed of boat A is v_a=9\ mi/hr

Speed of boat B is v_b=15\ mi/hr

Both are moving perpendicular to each other

Distance traveled by Boat A x_a=9t

Distance traveled by Boat B x_b=15t

Distance between them is given by Pythagoras theorem

\Rightarrow d^2=(x_a)^2+(x_b)^2\\\\\Rightarrow d^2=(9t)^2+(15t)^2\\\\\Rightarrow d=\sqrt{(9t)^2+(15t)^2}\\\\\Rightarrow d=\sqrt{81t^2+225t^2}\\\\\Rightarrow d=\sqrt{306t^2}\\\\\Rightarrow d=17.49t\approx 17.5t\ \text{miles}

Distance between them is 17.5t

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The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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