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3 years ago

Write a formula that converts x hours to y minutes

Mathematics

2 answers:

kykrilka [37]3 years ago

6 0

Formula:
y = 60x

1 hour = 60 minutes

Nikitich [7]3 years ago

4 0

Answer:

$x\text{ hours}\cdot\frac{60\text{ minutes}}{\text{Hour}}=y\text{ minutes}$

Step-by-step explanation:

We are asked to write a formula that converts x hours to y minutes.

We know that 1 hour equals 60 minutes.

$\text{Number of hours}\cdot\frac{60\text{ minutes}}{\text{Hour}}=\text{Number of minutes}$

Substitute the given values:

$x\text{ hours}\cdot\frac{60\text{ minutes}}{\text{Hour}}=y\text{ minutes}$

Therefore, the formula $x\text{ hours}\cdot\frac{60\text{ minutes}}{\text{Hour}}=y\text{ minutes}$ can be used to convert x hours to y minutes.

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How many times does 20 go into 74??

Dima020 [189]

About three times. Just divide 74 by 20 which is 3 but exactly it is 3.7.
hoped it helped

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3 years ago

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your bakery is in a large shopping center. The entrie shopping center is 4,500 square feet. Your bakery takes up only 1/10 of th

Kisachek [45]

So if the bakery is 1/10 of the shopping center's area, and the shopping center's area is 4500 feet, then the bakery's area is 1/10th of 4500 feet. 1/10th of 4500 feet is 450.

A more in-depth explanation:

A fraction is:
x/y
where y would equal how many parts you wanted to split a number into and x would equal how many parts you want to take.

So splitting 4500 into 10 parts and then taking one is basically 1/10 of 4500.

4500 in 10 parts, one would be 450.

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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.