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IRISSAK [1]
3 years ago
14

Complete this statement: 20 a x 2 + 25 a x + 15 a = 5 a 20 a x 2 + 25 a x + 15 a = 5 a

Mathematics
1 answer:
liubo4ka [24]3 years ago
4 0

Answer:

Step-by-step explanation:

(((22•5ax2) +  25ax) +  15a) -  5  = 0  

Step  2  :

Step  3  :

Pulling out like terms :

3.1     Pull out like factors :

  20ax2 + 25ax + 15a - 5  =  

 5 • (4ax2 + 5ax + 3a - 1)  

Equation at the end of step  3  :

 5 • (4ax2 + 5ax + 3a - 1)  = 0  

Step  4  :

Equations which are never true :  

4.1      Solve :    5   =  0

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Answer:

SA = 967.6 cm²

Step-by-step explanation:

Surface Area of cylinder = 2\pi r h + 2\pi r^2

<u><em>Where r = 7 cm, h = 15 cm</em></u>

=> SA = 2(3.14)(7)(15)+2(3.14)(7)²

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6 0
3 years ago
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How are the graphs of the functions f(x) = and g(x) = related?
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The x variable, that is being multiplied.
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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 .

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

n_1=155

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

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

                                     $=2 \times 0.085667$

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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.

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3 years ago
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8 0
3 years ago
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