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iren2701 [21]
3 years ago
14

Express f(x) =3\x²- 1 in partial fraction​

Mathematics
1 answer:
BARSIC [14]3 years ago
7 0

3/(<em>x</em> ² - 1) = 3/((<em>x</em> - 1) (<em>x</em> + 1)) = <em>a</em>/(<em>x</em> - 1) + <em>b</em>/(<em>x</em> + 1)

==>   3 = <em>a</em> (<em>x</em> + 1) + <em>b</em> (<em>x</em> - 1)

==>   3 = (<em>a</em> + <em>b</em>) <em>x</em> + <em>a</em> - <em>b</em>

==>   <em>a</em> + <em>b</em> = 3 and <em>a</em> - <em>b</em> = 0

==>   <em>a</em> = <em>b</em>

==>   <em>a</em> + <em>b</em> = 2<em>a</em> = 3   ==>   <em>a</em> = 2/3 and <em>b</em> = 2/3

So you have

3/(<em>x</em> ² - 1) = 2/(3(<em>x</em> - 1)) + 2/(3(<em>x</em> + 1))

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photoshop1234 [79]

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$33.60

Step-by-step explanation:

Since 20 percent is off, you only need to pay for 80% of the jeans. Therefore, you do 8/10 x 42 = 33.6

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Pentagon A'B'C'D' is the image of pentagon ABCDE under a dilation,
viva [34]

Answer:

scale factor = \frac{1}{3}

Step-by-step explanation:

Compare the ratio of corresponding sides, image to original.

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3 0
3 years ago
The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp
andreyandreev [35.5K]

Answer:

(a) X ~ N(\mu=63, \sigma^{2} = 13^{2}).

    \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the normal distribution is given by;

                      Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

So, the distribution of X ~ N(\mu=63, \sigma^{2} = 13^{2}).

Let \bar X = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

                      Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X \leq 61.4 mL)

  P(X < 62.8 mL) = P( \frac{X-\mu}{\sigma} < \frac{62.8-63}{13} ) = P(Z < -0.02) = 1 - P(Z \leq 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X \leq 61.4 mL) = P( \frac{X-\mu}{\sigma} \leq \frac{61.4-63}{13} ) = P(Z \leq -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < \bar X < 62.8 mL)

   P(61.4 mL < \bar X < 62.8 mL) = P(\bar X < 62.8 mL) - P(\bar X \leq 61.4 mL)

  P(\bar X < 62.8 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{62.8-63}{\frac{13}{\sqrt{43} } } ) = P(Z < -0.10) = 1 - P(Z \leq 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P(\bar X \leq 61.4 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{61.4-63}{\frac{13}{\sqrt{43} } } ) = P(Z \leq -0.81) = 1 - P(Z < 0.81)

                                                           = 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

4 0
3 years ago
In the figure above, quadrilateral ABCD is a parallelogram. Let x represent the measure of angle GBF, y represent the measure of
ira [324]
Answers:
measure angle x = 40°
measure angle y = 35°
measure angle z = 55°

Explanation:
Part (a): getting angle x:
In triangle BED, we have:
measure angle BED = 90°
measure angle BDE = 50°
Therefore:
measure angle DBE = 180 - (90+50) = 40°
Now, we have angle DBE and angle GBF vertically opposite angles.
This means that they are both equal. Therefore angle GBF = 40°
Since angle GBF is x, therefore:
x = 40°

Part (b): getting angle y:
We know that the sum of measures of angles on a straight line is 180.
This means that:
angle GBF + angle GBC + angle CBE = 180
We have:
angle GBF = 40°
angle GBC = 105°
angle CBE = y
Therefore:
40 + 105 + y = 180
y = 35°

Part (c): getting angle z:
In triangle BCE, we have:
measure angle BCE = z
measure angle BEC = 90°
measure angle CBE = 35°
Therefore:
z + 90 + 35 = 180
z = 55°

Hope this helps :)

8 0
3 years ago
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