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4 years ago

HELP PLEASE!!!!!!! This is due tomorrow. 20 points for this.

Mathematics

1 answer:

gogolik [260]4 years ago

5 0

Answer:

x=16.75 y=4.5

Step-by-step explanation:

first turn the first trapezoid sideways so the side that says 8 matches with the side that says 10. This is the proportion. 8:10. The bottom of the first trapezoid says x+2. This matches with the side that says 15. Because the side that says 15 is on the bigger trapezoid, multiply by ten. Then multiply the other trapezoid by 8. We get an equation that looks like. 8x+16=150 Simplify and 8x=134 divide and we get x=16.75. Do the same for the side with y. 8y-24=50. 8y=36 y=4.5. There you have it.

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Answer:

64cm^3

Step-by-step explanation:

4*4*4=64

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3 years ago

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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.