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3 years ago

The sum of the first 1 million primes is N. Without knowing N's value, one can determine that the ones'digit of N cannot be

1 answer:

3 0

All primes are odd, except two. Therefore the sum of the first million primes is one even number plus 999999 odd numbers. Since odd*odd = odd, the first million primes must be even + odd = odd.

An odd number obviously has an odd final digit, so the only option that can be firmly rules out is b (which is even).

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What is the value of this expresion <br> 4(-6^2+ 3) - 2-1/2

lesya [120]

Answer: -134.5

Step-by-step explanation:

4(-6^2+3) - 2 - 1/2 Exponents first

4(-36 +3) - 2 - 1/2 Now solve the parentheses

4(-33) -2 -1/2

-132- 2 -1/2

-134 - 1/2 = -134.5

7 0

3 years ago

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.