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sergiy2304 [10]
3 years ago
7

A boat travels with velocity vector (25, 25 StartRoot 3 EndRoot). What is the directional bearing of the boat?

Mathematics
2 answers:
GenaCL600 [577]3 years ago
7 0

Answer:

A on Egde

Step-by-step explanation:

i got it right

iogann1982 [59]3 years ago
6 0

Answer:

The directional earing is N 30° E

Step-by-step explanation:

The given velocity vector is (25, 25·√3), therefor, we have;

R = 25·i + 25·√3·j

The angle of inclination of the direction of the boat with the horizontal = Arctan (y/x)

Where;

(x, y) = The coordinate of the components of the velocity of the boat

Therefore;

Arctan (y/x)  =  Arctan (25·√3/25) = Arctan (√3) = 60° North of East

The directional bearing which is the bearing relative to the northern direction is given as follows;

The directional earing = 90° - 60° = N 30° E.

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<h2>Steps:</h2>

So for this, we will be completing the square to solve for m. Firstly, subtract 8 on both sides:

2m^2-16m=-8

Next, divide both sides by 2:

m^2-8m=-4

Next, we want to make the left side of the equation a perfect square. To find the constant of this perfect square, divide the m coefficient by 2, then square the quotient. In this case:

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Add 16 to both sides of the equation:

m^2-8m+16=12

Next, factor the left side:

(m-4)^2=12

Next, square root both sides of the equation:

m-4=\pm \sqrt{12}

Next, add 4 to both sides of the equation:

m=4\pm \sqrt{12}

Now, while this is your answer, you can further simplify the radical using the product rule of radicals:

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√12 = √4 × √3 = 2√3.

m=4\pm 2\sqrt{3}

<h2>Answer:</h2>

In exact form, your answer is m=4\pm \sqrt{12}\ \textsf{OR}\ m=4\pm 2\sqrt{3}

In approximate form, your answers are (rounded to the hundreths) m=7.46, 0.54

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student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
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Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

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= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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