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Elden [556K]
4 years ago
9

Michael and Michelle are at a restaurant. When leaving, they decide to leave an 18% tip. Michael states that in order to figure

out their total, they need to multiply their bill by 0.18 and add that product on to the bill. Michelle says that another way to figure their total is to multiply their bill by 1.18. Will Michelle's method work? Explain why or why not.
Use properties of operations to generate equivalent expressions.
​
Mathematics
2 answers:
Marta_Voda [28]4 years ago
5 0

Michelle’s method WILL work because we have to add an extra 18% to the already total (100) michaels way would work if the tip was apart of the bill, but it is extra . Hence, Michelle’s method being correct

IrinaVladis [17]4 years ago
4 0

Yes his method is correct

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6\frac{2}{3}

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3 years ago
The United States government used to make coins of many different values
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1. 50%

2. 3%

3. 20%

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$1 is same as 100 cents

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5 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}}}

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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

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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.

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