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sertanlavr [38]
2 years ago
7

What type of slope does this line have? y = 7

Mathematics
2 answers:
Softa [21]2 years ago
5 0

Answer:

<h2>0</h2>

Step-by-step explanation:

The slope of this line is 0 and the intercept is 7.

Gemiola [76]2 years ago
5 0

Answer:

the slope is 0 and the y-intercept is -7 then i think its an undefined slope

Step-by-step explanation:

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Your class wants to know to most typical age of student in your class. It collects the ages of all students and the teacher. Whi
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Median.

Step-by-step explanation:

Assuming the teacher is not of similar age to the students, their higher age would pull the mean higher than if the teacher were not included, where  the median would not be pulled by this extreme value.

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110 is 20% of what number
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Your answer is probably:
22
6 0
3 years ago
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Please help, idek what to do and it's a test!!!! I'll give 25 points!!!!
Schach [20]

Answer:

Ok so what you are suppose to do is find the area

Step-by-step explanation:

Which here is an explanation of how to find area (The simplest (and most commonly used) area calculations are for squares and rectangles. To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.)

8 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}}}

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.

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