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3 years ago

In ΔGHI, h = 510 inches, ∠G=25° and ∠H=122°. Find the length of g, to the nearest inch.

Mathematics

1 answer:

Licemer1 [7]3 years ago

8 0

9514 1404 393

Answer:

254 in

Step-by-step explanation:

The Law of Sines tells you ...

g/sin(G) = h/sin(H)

g = h(sin(G)/sin(H)) = 510 in(0.42262/0.84805)

g ≈ 254.15

The length of g is about 254 inches.

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What is the length of bc in the right triangle below

worty [1.4K]

To solve for this, we need to use Pythagorean Theorum.
Pythagorean Theorum is when we add the squares of the legs (A & B) to solve for the hypotenuse (C) or the side that's across from the 90 degree angle.
The formula for Pythagorean Theorum is A^2 + B^2 = C^2.

We have our two leg lengths, 15 and 36 (A & B).
Square 15 and 36.
15^2 = 225
36^2 = 1,296

Now that we have our side lengths, let's add them up.
225 + 1,296 = 1,521.

Now, to solve for C, we need to square root our sum.

The square root of 1,521 is 39.

Your hypotenuse is 39.

Your answer is C.) 39.

I hope this helps!

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3 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$