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Arturiano [62]
3 years ago
13

Identify the 33rd term of the arithmetic sequence 9, 7 1/2, 6.

Mathematics
1 answer:
sertanlavr [38]3 years ago
7 0
The formula of arithmetic sequence is expressed in the following manner:
<span>an = a1 + d*(n-1) where d is the difference and n is an integer.
substituting, 
7  = 9 + d*(2-1)
d= -1.5

hence, 
a33 = 9 - 1.5*(33-1)
a33 = -39</span>
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What is NOT an example of a short term debt?
polet [3.4K]

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C

Step-by-step explanation:

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Simplify the expression -2x^2(4x − 3).<br> -8x3 − 6x2<br> -8x3 + 6x2<br> 8x3 − 6x2<br> 8x3 + 6x2
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To simplify the expression we are going to use the distributive property of multiplication: a(b+c)=ab+ac.
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8 0
3 years ago
If 4x+5=0 what is the value of x​
netineya [11]

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

                                     $=2 \times 0.085667$

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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.

8 0
3 years ago
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