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Colt1911
16 days ago
8

A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each r

ung in 20cm and the bottom rung touches the water. The tide rises at a rate of 10cm an hour. When will the water reach the fifth rung
Mathematics
1 answer:
Leona [9.2K]16 days ago
3 0

Response:

The answer to the inquiry is 8 hours.

Step-by-step breakdown:

Information

Ladder length = 200 cm

Distance between rungs = 20 cm

Tide rise rate = 10 cm/h

Fifth rung =?

Procedure

1.- Determine the total height the tide must reach

Height = 20 cm x 4

                 = 80 cm   as the first rung is touching the water.

2.- Calculate the time needed

Rate = distance / time

-Solve for time

Time = distance / rate

-Substitute values

Time = 80 cm / 10 cm/h

-Final outcome

Time = 8 hours.

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lawyer [9240]
The distance AB measures 185.77 feet.
Step-by-step explanation:
According to the attached figure, in triangle ABC,
∠B = 103.2°, ∠C = 14.4° and side BC = 661 feet.
We need to calculate AB's measure.
Since ∠A + ∠B + ∠C = 180°
∠A + 103.2 + 14.4 = 180
∠A + 117.6 = 180
∠A = 180 - 117.6
= 62.4°
Utilizing the sine rule in the triangle ABC,
AB =
= 185.77 feet.
Thus, AB = 185.77 feet is the solution.
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9 days ago
Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th
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Answer:

Step-by-step reasoning:

(a)

To have an accepted bid, it needs to exceed $10,000. Let bid x be a continuous random variable uniformly distributed between

$10,000 and $15,000

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The provided bidding range is [$10,000,$12,000]. The probability is determined as follows,

\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}

=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4

(b)  The accepted bidding range is [$10,000,$15,000], where b = $15,000 and a =$10,000. The given bidding range is [$10,000,$14,000].

\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)

=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8

(c)

The optimal amount to bid for maximizing the probability of acquiring the property is calculated as,  

The accepted bidding range is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The provided bidding range is [$10,000,$15,000].

\begin{array}{c}\\f\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\\\\{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\\\\{\rm{ = 1}}\\\end{array}

(d)  If you know someone willing to pay you $16,000 for the property, would you still consider bidding less than the amount mentioned in part (c)? Why or why not?

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5 days ago
Tucker was asked to solve the equation 5x + 3 = 6x + 1. He did not know if his first step should be to add 5 negative x-tiles, o
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It doesn't matter what step is taken first. Either choice will result in zero pairs being created on both sides, which helps isolate the variable x. Whether you start with the x-tiles or the unit tile, the solution will be the same.
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1 month ago
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Could you please rephrase the question? I'm having trouble understanding it.
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25 days ago
For every 10 yards on a football field, there is a boldly marked line labeled with the amount of yards. Each of those lines is p
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Response:

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