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Partial fraction decomposition 20x+9/25x^2+20x+4

ozzi

Step-by-step explanation:

We have

$\frac{20x + 9}{25 {x}^{2} + 20x + 4 }$

Let's factor the denomiator first,

the denomaitor is a perfect square so we get

$\frac{20x + 9}{(5x + 2) {}^{2} }$

Now, we must think of two fractions that

$\frac{a}{(5x + 2) {}^{2} } + \frac{b}{5x + 2}$

We use a perfect square term for one fraction, then a linear one for the next, because if we set both of the denomiator to the same factor, we would get a inconsistent system.

So right now, we have

$\frac{a}{(5x + 2) { }^{2} } + \frac{b}{5x + 2} = \frac{20x +9 }{25 {x}^{2} + 20x + 4 }$

$a + (5x + 2)b = 20x + 9$

$5b (x)= 20$

$a + 2b = 9$

$b = 4$

so that means that a is

$a + (2)(4) = 9$

$a + 8 = 9$

$a = 1$

So our equation is

$\frac{1}{(5x + 2) {}^{2} } + \frac{4}{5x + 2}$

6 0

2 years ago

The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2617 and standard deviation

Eduardwww [97]

Answer:

a) N(2617, 586)

b) 0.3613 = 36.13% probability that the customer consumes less than 2409 calories.

c) 0.4013 = 40.13% of the customers consume over 2764 calories

d) 3981 calories.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 2617, \sigma = 586$

a. What is the distribution of X?

Here we first place the mean, then the standard deviation.

N(2617, 586)

b. Find the probability that the customer consumes less than 2409 calories.

This is the pvalue of Z when X = 2409. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2409 - 2617}{586}$

$Z = -0.355$

$Z = -0.355$ has a pvalue of 0.3613

0.3613 = 36.13% probability that the customer consumes less than 2409 calories.

c. What proportion of the customers consume over 2764 calories?

This is 1 subtracted by the pvalue of Z when X = 2764. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2764 - 2617}{586}$

$Z = 0.25$

$Z = 0.25$ has a pvalue of 0.5987

1 - 0.5987 = 0.4013

0.4013 = 40.13% of the customers consume over 2764 calories

d. The Piggy award will given out to the 1% of customers who consume the most calories. What is the fewest number of calories a person must consume to receive the Piggy award?

Top 1%, so the 100-1 = 99th percentile.

The 99th percentile is the value of X when Z has a pvalue of 0.99. So it is X when Z = 2.327. So

$Z = \frac{X - \mu}{\sigma}$

$2.327 = \frac{X - 2617}{586}$