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Which method correctly solves the equation using the multiplication property of equality and the reciprocal of One-third? One-th

Airida [17]

Answer:

Step-by-step explanation:

Given the equation $\frac{1}{3}(x+9) = -12$

Step 1;

Expand the bracket at the right hand side of the equation to have:

$\frac{1}{3}x +\frac{1}{3}(9) = -12\\\frac{1}{3}x+3=-12\\subtracting\ 3\ from\ both\ sides\\\frac{1}{3}x+3-3=-12-3\\\frac{1}{3}x=-15\\$

Taking the reciprocal of both sides:

$\frac{3}{x} = \frac{-1}{15} \\ cross\ multiplying\\-x =45\\x=-45$

4 0

4 years ago

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Anton took out a 20-year loan for $90,000 at an APR of 10.5% compounded monthly, approximately what would be the total cost of h

snow_lady [41]

The answer would be $159,661.69. i think

5 0

3 years ago

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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.