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trasher [3.6K]
2 years ago
12

DON’T SEND A LINK! WHAT IS THE SLOPE OF THIS LINE THE SLOPE IS NOT 4/1 OR 20/5

Mathematics
2 answers:
katrin [286]2 years ago
8 0

Answer:

slope = -4

Step-by-step explanation:

m=y2-y1/x2-x1

20-0/0-5

20/-5

= -4

aleksandrvk [35]2 years ago
7 0

Answer:

25/5

Step-by-step explanation:

i think

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Answer:

True.

Step-by-step explanation:

Remember the vertical line test. Anywhere on that graph, you can draw a vertical line, and it will not touch the function two times. Because of this, it is true.

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Amelia drives 10 miles in 20 minutes. If she drove three hours in total at the same rate, how far did she go? subquestion - how
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Answer:

Amelia drove 90 miles in 3 hours.

Step-by-step explanation:

she drove 10 miles in 20 minutes

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Suppose that in a random sample of 300 employed Americans, there are 57 individuals who say that they would fire their boss if t
otez555 [7]

Answer:

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 300, \pi = \frac{57}{300} = 0.19

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 - 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.1456

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 + 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.2344

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

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