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2 years ago

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Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes

and Boat 1 now sails at 9 knots. At 3 pm, the wind increases for Boat 2 and it now sails 7 knots faster. How fast (in knots) is the distance between the two ships changing at 5 pm. (Note: 1 knot is a speed of 1 nautical mile per hour.)

1 answer:

Ivan2 years ago

6 0

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2 $\times$ 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 $\times$ 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled $\triangle AOB$ .

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units$

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = $\dfrac{Total\ distance}{Total\ time}$

$\Rightarrow \dfrac{74.28}{5} = 14.86\ knots$

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a) The random variable is the mean yearly rainfall in Sydney, Australia.

b) There is a 29.46% probability that the yearly rainfall is less than 100 mm.

c) There is a 6.81% probability that the yearly rainfall is more than 240mm.

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Problems of normally distributed samples can be solved using the z-score formula.

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After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

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The mean yearly rainfall in Sydney, Australia, is about 137 mm and the standard deviation is about 69 mm, so $\mu = 137, \sigma = 69$ .

a.) State the random variable.

The random variable is the mean yearly rainfall in Sydney, Australia.

b.) Find the probability that the yearly rainfall is less than 100 mm.

This is the pvalue of Z when $X = 100$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{100 - 137}{69}$

$Z = -0.54$

$Z = -0.54$ has a pvalue of 0.2946.

This means that there is a 29.46% probability that the yearly rainfall is less than 100 mm.

c.) Find the probability that the yearly rainfall is more than 240 mm.

This is 1 subtracted by the pvalue of Z when $X = 240$ .

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{240 - 137}{69}$

$Z = 1.49$

$Z = 1.49$ has a pvalue of 0.9319.

So, there is a 1-0.9319 = 0.0681 = 6.81% probability that the yearly rainfall is more than 240mm.

d.) Find the probability that the yearly rainfall is between 140 and 250 mm.

This is the pvalue of Z when $X = 250$ subtracted by the pvalue of Z when $X = 140$ .

For X = 250

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{250 - 137}{69}$

$Z = 1.64$

$Z = 1.64$ has a pvalue of 0.9495

For X = 140

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$Z = \frac{140 - 137}{69}$

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A probability is unusually low when there is a lesser than 5% of it happening.

Since the probability of a year having a rainfall of less than 100mm is 29.46%, as we found in b), it would not be an unusually dry year.

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This rainfall is the 10th percentile of rainfall.

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$Z = \frac{X - \mu}{\sigma}$

$-1.285 = \frac{X - 137}{69}$

$X - 137 = -88.665$

$X = 48.335$

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