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3 years ago

Solve the right triangle ABC with right angle C if B = 30° and c = 10.

Mathematics

1 answer:

mart [117]3 years ago

6 0

Step-by-step explanation:

Answer:

Last choice is correct.

Step-by-step explanation:

Given that ∠C=90°, ∠B=30°, c=10

We know that sum of all angles in a triangle is 180 degree.

∠A+∠B+∠C=180°

∠A+∠30°+∠90°=180°

∠A+∠120°=180°

∠A=60°

Now we can use sin or cos whichever applicable to find missing sides

2b=10

b=5

Similarly

a=8.6602

Hence last choice is correct.

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Goryan [66]

Answer:

6. 10

Step-by-step explanation:

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3 years ago

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MariettaO [177]

Answer:

y = -1/2x + -3

Step-by-step explanation:

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3 years ago

John ran 1 1/4 laps in 2 minutes. How many laps did John run on average per minutes?

alukav5142 [94]

Answer:

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Step-by-step explanation:

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4 years ago

Consider a bettering game where you bet $10 and have a probability of 0.45 of getting $20 back ($10 more than you started with)

Marina CMI [18]

Answer:

(a)

$f(x) = P(X=x) = \begin{Bmatrix} 0.45 \,\,\, \text{for} \,\,\, x = 10 \\ 0.55 \,\,\, \text{for} \,\,\, x = -10 \end{matrix}$

(b)

-1

(c)

Step-by-step explanation:

(a)

Your random variable will have two possible values, 30 and 0, one of them with a probability of 0.45 and the other one with a probability of 0.55. Therefore it looks like this.

$f(x) = P(X=x) = \begin{Bmatrix} 0.45 \,\,\, \text{for} \,\,\, x = 10 \\ 0.55 \,\,\, \text{for} \,\,\, x = -10 \end{matrix}$

(b)

The expected value of this PMF would be

$E[X] = -10*0.55+10*0.45= -1$ therefore on average you will have a dollar less.

(c)

For this one, if you play 20 times and your initial amount is 50$ then you have to compute the following expectation.

$E[50+20*X] = 50+20*E[X] = 50-20 = 30$

7 0

3 years ago

Determine the gravitational force of attraction between two 3.5 kg bowling balls whose

Ipatiy [6.2K]

Answer:

$F_G=2.04\ .\ 10^{-10}\ Nw$

Option J

Step-by-step explanation:

Newton's Universal Law Of Gravitation

Newton discovered that all objects attract each other with a force of gravitational attraction which is proportional to their masses and inversely proportional to the square of the distance between them.

$\displaystyle F_G=G\frac{m_1m_2}{d^2}\ ,\ G=6.673\ .\ 10^{-11} N m^2/kg^2$