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2 years ago

Question is 15 points

Mathematics

1 answer:

Illusion [34]2 years ago

5 0

In every right triangle, legs are hypothenuse are linked by the following relationship: the sine of an angle is given by the ratio of the opposite leg and the hypothenuse, while the cosine of an angle is given by the ratio of the adjacent leg and the hypothenuse.

In this case, the hypothenuse is 391 units long, while the 267-units-leg is adjacent to the angle you're interested in.

For this reason, you have

$\cos(A) = \cfrac{267}{391} \approx 0.68286445$

To find the value of the angle, simply invert this formula:

$A = \arccos(0.68286445) \approx 46.93$

If we need to approximate this value to the nearest degree, we can observe that 46.93 lies between 46 and 47. So, we have to choose the closest one.

46 is 0.93 units far, while 47 is only 0.07 units far, so the answer is 47.

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3 years ago

What is the value of x? A.39 B.41 C.26 D.17

mojhsa [17]

Answer:

The answer is B.

Step-by-step explanation:

You have to use Pythogorean Theorem, c² = a² + b² where c is the hypotenuse and a,b is the side lengths :

${c}^{2} = {a}^{2} + {b}^{2}$

Let a = 40,

Let b = 9,

${c}^{2} = {40}^{2} + {9}^{2}$

${c}^{2} = 1600 + 81$

${c}^{2} = 1681$

$c = \sqrt{1681}$

$c = 41 \: units$

*Take note, hypotenuse will always be the longest length among the 3 lengths.

*Pythogorean Theorem can only applied in a right-angle triangle.

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2 years ago

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 b

jeka94

Answer:

Step-by-step explanation:

From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .

So; as each place is fitted with either 0 or 1

$\mathbf{|R|= 2^{10}}$

Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0

Now;

if a 0 bit and a 1 bit are equally likely

The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;

$\mathbf{P(E) = \dfrac{|E|}{|R|}}$