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3 years ago

Prove (1+sin/csc-cot)-(1-sin/csc+cot)=2(1+cot)

1 answer:

3 0

$\dfrac{1+sin}{csc-cot}-\dfrac{1-sin}{csc+cot}=2(1+cot)\\\\\\\text{Proof from LHS }\rightarrow \text{ RHS}\\\\\dfrac{1+sin}{csc-cot}\bigg(\dfrac{csc+cot}{csc+cot}\bigg)-\dfrac{1-sin}{csc+cot}\bigg(\dfrac{csc-cot}{csc-cot}\bigg)\\\\\\\dfrac{csc+cot+sin csc+sincot}{csc^2-cot^2}+\dfrac{-csc+cot+sincsc-sincot}{csc^2-cot^2}\\\\\\\dfrac{2cot+2sincsc}{csc^2-cot^2}\\\\\\\dfrac{2(cot+sincsc)}{1}\\\\\\2\bigg(\dfrac{cos}{sin}+\dfrac{sin\cdot1}{sin}\bigg)\\\\\\2(cot+1)\\\\\\2(1+cot)=2(1+cot)$

LHS = RHS $\checkmark$

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A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

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

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

3 years ago

c. Using a standard deck of 52 cards, the probability of selecting a 4 of diamonds or a 4 of hearts is an example of a mutually

dybincka [34]

Answer:

True

Step-by-step explanation:

If two events X and Y are mutually exclusive,

Then,

P(X∪Y) = P(X) + P(Y)

Let A represents the event of a diamond card and B represent the event of a heart card,

We know that,

In a deck of 52 cards there are 4 suit ( 13 Club cards, 13 heart cards, 13 diamond cards and 13 Spade cards )

That is, those cards which are heart can not be diamond card,

⇒ A ∩ B = ∅

⇒ P(A∩B) = 0

Since, P(A∪B) = P(A) + P(B) - P(A∩B)

⇒ P(A∪B) = P(A) + P(B)

By the above statement,

Events A and B are mutually exclusive,

Hence, the probability of selecting a 4 of diamonds or a 4 of hearts is an example of a mutually exclusive event is a true statement.

4 0

4 years ago

Mark filed a claim after his car was rear-ended by another driver. Which

Fofino [41]

Answer: A. He would not have to pay

Step-by-step explanation:

3 0

3 years ago

What is the equation of the line whose y-intercept is 3 and slope is 1?

AlexFokin [52]

Answer:

B. y = x + 3

Step-by-step explanation:

Let m represent the slope of the line.

Let c represent the y-intercept of the line.

Then the equation of the line with slope m and y-intercept c is given by $\[y=mx+c\]
$

In this case, the value of m is 1 and the value of c is 3.

Substituting the given values in the equation:

$\[y=x+3\]
$

Among the given options, option B represents the correct equation of the line.

3 0

3 years ago

Step by step If C(m)=0.50m + 30 represents the cost of renting a car, how many miles were driven if the cost is $130

Likurg_2 [28]

Answer:

200 miles were driven

Step-by-step explanation:

We know that $C (m) = 0.50m + 30$ represents the cost of renting a car

Where m represents the number of miles driven

If we know that the cost was $ 130 then we can equal C(m) to 130 and solve for m.

$C(m) =0.50m + 30=130$

$0.50m + 30=130$

$0.50m=130-30$

$0.50m=100$

$m=\frac{100}{0.50}$

$m=200\ miles$

7 0

4 years ago

Prove (1+sin/csc-cot)-(1-sin/csc+cot)=2(1+cot)​

Prove (1+sin/csc-cot)-(1-sin/csc+cot)=2(1+cot)