If t = 18 and r = 12, find S. Round to the nearest tenth.

Mathematics

2 answers:

KatRina [158]2 years ago

7 0

Answer:

13.4 or √180

Step-by-step explanation:

Pythagorean theorem states

a² + b² = c²

C^2 is the longest side. The corner of the right angle points to it. The other sides are your a and b squared. They are called the legs or Opposite and Adjacent sides.

choli [55]2 years ago

3 0

Answer: S ≈ 48.2°

Step-by-step explanation:

We can use trigonometry functions to solve.

Looking at angle S, t = 18 is the hypotenuse and r = 12 is the adjacent side. This means we can use the cosine function.

$\displaystyle cos(S) = \frac{\text{adjacent side}}{\text{hypotenuse }}=\frac{12}{18}$

$\displaystyle cos^{-1} (cos(S)) = (\frac{\text{adjacent side}}{\text{hypotenuse }}=\frac{12}{18})cos^{-1}$

$S = 48.189685...\\S \;$\approx$\; 48.2^{\circ}$

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svetoff [14.1K]

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3 0

3 years ago

Please help (-0.5)/(-0.5 ) -50/(-0.5 )

Kipish [7]

If you would like to solve (- 0.5) / (- 0.5) and - 50 / (- 0.5), you can do this using the following few steps:

(- 0.5) / (- 0.5) = 0.5 / 0.5 = 1

- 50 / (- 0.5) = 50 / 0.5 = 50 / (5/10) = 50 / (1/2) = 50 * 2/1 = 100

The correct results would be: (- 0.5) / (- 0.5) = 1 and - 50 / (- 0.5) = 100.

6 0

3 years ago

A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received wit

m_a_m_a [10]

Answer:

The probability that the message will be wrong when decoded is 0.05792

Step-by-step explanation:

Consider the provided information.

To reduce the chance or error, we transmit 00000 instead of 0 and 11111 instead of 1.

We have 5 bits, message will be corrupt if at least 3 bits are incorrect for the same block.

The digit transmitted is incorrectly received with probability p = 0.2

The probability of receiving a digit correctly is q = 1 - 0.2 = 0.8

We want the probability that the message will be wrong when decoded.

This can be written as:

$P(X\geq3) =P(X=3)+P(X=4)+P(X=5)\\P(X\geq3) =\frac{5!}{3!2!}(0.2)^3(0.8)^{2}+\frac{5!}{4!1!}(0.2)^4(0.8)^{1}+\frac{5!}{5!}(0.2)^5(0.8)^0\\P(X\geq3) =0.05792$