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3 years ago

What is the decimal for 1/3

Mathematics

2 answers:

monitta3 years ago

7 0

The decimal for 1/3 is 0.333 repeat.

yarga [219]3 years ago

5 0

You can find the decimal form of any fraction by doing the math that is defined by the fraction. The fraction bar that divides the numerator and denominator means "divided by". 1/3 literally reads 1 divided by 3. When you do the division...

1 ÷ 3 = .3333 repeating

.33 rounded to the nearest hundredth

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It's question B as I've done A please help

Hatshy [7]

Its easy.use cos rule.

4 0

3 years ago

4. Joseph is making cookies. The recipe requires 1/2 cups of flour and 1 teaspoons of sugar.

oksian1 [2.3K]

Answer:

He will use 4.

Step-by-step explanation:

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2 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.