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SVEN [57.7K]
2 years ago
5

Help me im so sad determine the values of X for which f(x) is defined ​

Mathematics
1 answer:
eimsori [14]2 years ago
4 0

Answer:

f(x) =  \frac{(2x - 1)(3x - 2)}{9x - 4}  \\  open \: brackets \\  =  \frac{(6 {x}^{2} - 4x - 3x + 2) }{9x - 4}  \\  =  \frac{6 {x}^{2} - 7x + 2 }{9x - 4}  \\ let \: x \: be \: 1 \\ f(1) =  \frac{(6 - 7 + 2)}{(9 - 4)}  \\  =  \frac{1}{5}

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+ First we have y=0 and x=m or 0=a*m+b or b= -a*m
+ Then we have x=0, y=n or n=a*0+b or b=n
And if m≠0 we have a= -n/m
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3 years ago
Please answer 100 points<br> Divide. 6 from 3/5 <br> 10<br> 30<br> 20<br> 15
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Answer:

Step-by-step explanation:

What is 6 divided by 3/56 ÷ 3 5. The trick to working out 6 divided by 3/5 is similar to the method we use to work out dividing a fraction by a whole number. ... 6 x 5 3 = 30 3. So, the answer to the question "what is 6 divided by 3/5?" ... 30 3. ... 10 1. ... 10 1 = 10.

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2 years ago
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3 years ago
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An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

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3 years ago
3(x-2)5 simplified and tiles
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Answer:

Step-by-step explanation:

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2 years ago
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