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densk [106]
1 year ago
9

I need helpppppp asap

Mathematics
1 answer:
user100 [1]1 year ago
4 0

Answer:

graph A

Step-by-step explanation:

When looking at a graph, there are two different axes. The vertical values--marked by the center up/down line--are "y-values"; and this is called the "y-axis"

The horizontal values--marked by the left/right line--are "x-values"; and this is called the "x-axis"

For the x-axis, values to the left side of the origin (the place where the y-axis and x-axis intercept) are smaller than 0--they are all negative values.

Values to the right side of the origin are positive--greater than 0.

For the y-axis, positive numbers are on the top half [once again, the midpoint / 0 is where the two lines are both = to 0; the origin] and negative numbers are on the bottom half.

Ordered pairs (points) are written as (x,y)

(x-value, y-value)

We are looking for a graph that decreases (along the y-axis), hits a point below the origin, and goes flat/stays constant.

When a graph is decreasing (note: we read graphs from left to right), the line of the graph is slanted downwards (it looks like a line going down).

So, if we look at the graphs, we can see Graph A descending, crossing the y-axis {crossing the middle line /vertical line / y-axis} at a value of -7, and then staying constant (it is no longer increasing or decreasing because the y-values stay the same)

hope this helps!!

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

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In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

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We know that the confidence interval is given by:

12.6 < \mu < 14.8

For this case we can find the margin of error like this:

ME= \frac{14.8-12.6}{2}= 1.1

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

12.6 +1.1=13.7

14.8-1.1=13.7

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

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3 years ago
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