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sergey [27]
3 years ago
11

What is the of A in a partial fraction decomposition below?

Mathematics
1 answer:
denis-greek [22]3 years ago
8 0

Answer:

3

Step-by-step explanation:

(8x + 28) / ((x − 1) (x + 2) (x + 3)) = A / (x − 1) + B / (x + 2) + C / (x + 3)

Multiply both sides by (x − 1) (x + 2) (x + 3).

8x + 28 = A (x + 2) (x + 3) + B (x − 1) (x + 3) + C (x − 1) (x + 2)

If x = 1:

8 + 28 = A (3) (4)

36 = 12A

A = 3

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Step-by-step explanation:

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Please help simplify this...
juin [17]
Answer is 36.
First use the negative exponent rule. When a number is put to the negative power it becomes a fraction
4/3^2

Now 3 to the second power is 9
4/1/9 (4 over 1/9)
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An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so
alexgriva [62]

Solution :

Let p_1 and p_2  represents the proportions of the seeds which germinate among the seeds planted in the soil containing 3\% and 5\% mushroom compost by weight respectively.

To test the null hypothesis H_0: p_1=p_2 against the alternate hypothesis  H_1:p_1 \neq p_2 .

Let \hat p_1, \hat p_2 denotes the respective sample proportions and the n_1, n_2 represents the sample size respectively.

$\hat p_1 = \frac{74}{155} = 0.477419

n_1=155

$p_2=\frac{86}{155}=0.554839

n_2=155

The test statistic can be written as :

$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}

which under H_0  follows the standard normal distribution.

We reject H_0 at 0.05 level of significance, if the P-value or if |z_{obs}|>Z_{0.025}

Now, the value of the test statistics = -1.368928

The critical value = \pm 1.959964

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

                                     = 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$, so we fail to reject H_0 at 0.05 level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the \text{proportion} of the seeds that \text{germinate differs} with the percent of the \text{mushroom compost} in the soil.

8 0
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