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bezimeni [28]
3 years ago
13

An insulated cooler holds 184 ounces of lemonade. How many full 12.4 ounce glasses of lemonade can be poured from the cooler?

Mathematics
2 answers:
hichkok12 [17]3 years ago
7 0
Yes, you are right divide 184/12.4 = 14.8 . In this case you will round up  because the number after the decimal point is more than five. Your answer will be 15 glasses.   
krok68 [10]3 years ago
6 0
You're right, that's the required operation.
\frac{184}{12.4}=15.08
Because they want to know how many glasses can be full, the answer is 15.
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Rice weighing 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?
QveST [7]

Answer:

The answer is 15 ounces

Step-by-step explanation:

33/4 ÷ 4 pounds.

(4 × 3 + 3)/4 ÷ 4 pounds.

15/4 ÷ 4 pounds.

15/4 × 1/4 pounds.

15/16 pounds.

Now we know that, 1 pound = 16 ounces.

So, 15/16 pounds = 15/16 × 16 ounces = 15 ounces.

Thus, The answer is 15 ounces

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frosja888 [35]

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Step-by-step explanation:

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It went down 5.5°C a minute.

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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 = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

                                     $=2 \times 0.085667$

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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.

8 0
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