Which expression is equivalent to -6(3x-2/3) -2x+9 -3x-6 2/3 -18x+12 -18x+4
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Answer:
Null Hypothesis, : p
45%
Alternate Hypothesis, : p < 45%
Step-by-step explanation:
We are given that the proportion of women in similar sales positions across the country in 2004 is less than 45%.
The collected random sample of size 50 showed that only 18 were women.
<u><em>Let p = proportion of women in similar sales positions across the country in 2004</em></u>
So, Null Hypothesis, : p
45%
Alternate Hypothesis, : p < 45%
Here, <u><em>null hypothesis states that</em></u> the proportion of women in similar sales positions across the country in 2004 is more than or equal to 45%.
On the other hand, <u><em>alternate hypothesis states that</em></u> the proportion of women in similar sales positions across the country in 2004 is less than 45%.
The test statistics that would be used here is One-sample z proportion test statistics, i.e;
T.S. = ~ N(0,1)
Hence, the above hypothesis is appropriate for the given situation.
Part A. The technique on how to find the equation that only applies to point D and E, is to create a line or curve that only includes two of these points. In this case, I created a random parabola that isolates points C and F from the rest of the points. First, we have to find the equation of the parabola through its general forms:
(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)
For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:
(-2 - 0)² = +4a(4 - 3)
4a = 4
So, the equation of the parabola is:
x² = 4(y - 3)
x² = 4y - 12
Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4
Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).
3 ? 7(4) − 4
3 ? 24
3 < 24
Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.
(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)
For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:
(-2 - 0)² = +4a(4 - 3)
4a = 4
So, the equation of the parabola is:
x² = 4(y - 3)
x² = 4y - 12
Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4
Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).
3 ? 7(4) − 4
3 ? 24
3 < 24
Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.