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3 years ago

Which values represent solutions to the equation? Tan(3Pi/4-2x)=-1, where x = [0,pi]

Mathematics

2 answers:

Juli2301 [7.4K]3 years ago

4 0

0 is the a answer definately

Lorico [155]3 years ago

3 0

Answer:{0, pi/2, pi}

Step-by-step explanation:

Got it right

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If ,f(x)=3x-6/x-2 what is the average rate of change of f(x) over the interval [6, 8]?

sukhopar [10]

I hope this helps you

f (x)= 3 (x-2)/(x-2)

f (x)= 3

f(6)= 3

f (8)= 3

4 0

4 years ago

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Determine the horizontal, vertical, and slant asymptotes: y=x2+2x-3/x-7

densk [106]

Answer:

vertical asymptote: x = 7
slant asymptote: y = x+9

Step-by-step explanation:

The vertical asymptotes are found where a denominator factor is zero (and there is no corresponding numerator factor to cancel it). Here, that is at x = 7.

There is no horizontal asymptote because the numerator is of higher degree than the denominator.

When you divide the numerator by the denominator, you get ...

y = (x +9) +60/(x -7)

Then when x gets large, the behavior is governed by the terms not having a denominator: y = x +9. This is the equation of the slant asymptote.

3 0

3 years ago

What are some units of measurement we could use for a postage<br> stamp?<br> (Please help me)

nordsb [41]

I believe the answer is cubic centimeter

8 0

3 years ago

Read 2 more answers

HELP ASAP

Alex787 [66]

<span>v(x)=(s/t)
= (3x - 6) / (-3x+6)
= [3(x-2)] / [-3(x-2)] --> 3 is factored out
= 1/-1 </span>---> common terms are cancelled out.
= -1 ---> This is the simplified formula.

To find the domain, we equate the denominator to 0.
-3x+6 = 0
3x = 6
x = 2
Domain: all values except 2.

w(x)=(t/s)(x)
= (-3x+6)x / (3x-6)
= [-3x(x-2)] / [3(x-2)] --> 3 is factored out
= -x --> The common terms are cancelled out. This is the simplified formula.

Solving for domain:
3x-6 = 0
3x = 6
x = 2
Domain: all values except 2.

4 0

3 years ago

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