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GenaCL600 [577]
2 years ago
14

PLEASE HELP! SHOW WORK AND/OR EXPLANATION PLEASE.

Mathematics
1 answer:
Gre4nikov [31]2 years ago
3 0

Answer:

1.  x = 55°

2.  x = 11°

3.  x = 30°

4.  x = 19°

Step-by-step explanation:

<u>Trigonometric Identities</u>

\sin(x)=\cos(90^{\circ}-x)

\cos(x)=\sin(90^{\circ}-x)

<u>Question 1</u>

\begin{aligned}\implies \sin(35) & =\cos(90-35)\\ & = \cos(55)\end{aligned}

\implies x=55^{\circ}

<u>Question 2</u>

\begin{aligned}\implies \cos(79) & =\sin(90-79)\\ & = \sin(11)\end{aligned}

\implies x=11^{\circ}

<u>Question 3</u>

\begin{aligned}\cos(x)& =\sin(2x)\\\implies \cos(x)& =\sin(90^{\circ}-x)\\\\\implies 2x & = 90-x\\3x & = 90\\x & = 30^{\circ}\end{aligned}

<u>Question 4</u>

\begin{aligned}\sin(x+4)& =\cos(4x-9)\\\implies \sin(\theta)& =\cos(90^{\circ}-\theta)\\\\\implies x+4 & = 90-(4x-9)\\x+4 & = 90-4x+9\\5x & = 95\\x & = 19^{\circ}\end{aligned}

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Using the binomial distribution, it is found that there is a 0.8295 = 82.95% probability that at least 5 received a busy signal.

<h3>What is the binomial distribution formula?</h3>

The formula is:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 0.54% of the calls receive a busy signal, hence  p = 0.0054.
  • A sample of 1300 callers is taken, hence n = 1300.

The probability that at least 5 received a busy signal is given by:

P(X \geq 5) = 1 - P(X < 5)

In which:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

Then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{1300,0}.(0.0054)^{0}.(0.9946)^{1300} = 0.0009

P(X = 1) = C_{1300,1}.(0.0054)^{1}.(0.9946)^{1299} = 0.0062

P(X = 2) = C_{1300,2}.(0.0054)^{2}.(0.9946)^{1298} = 0.0218

P(X = 3) = C_{1300,3}.(0.0054)^{3}.(0.9946)^{1297} = 0.0513

P(X = 4) = C_{1300,4}.(0.0054)^{4}.(0.9946)^{1296} = 0.0903

Then:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0009 + 0.0062 + 0.0218 + 0.0513 + 0.0903 = 0.1705.

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.1705 = 0.8295

0.8295 = 82.95% probability that at least 5 received a busy signal.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0
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