Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 2−x and y = 4x + 3 intersect are the s
1 answer:
Answer:
Step-by-step explanation:
Part A:
The solution of a system is not just the x coordinates; it is the whole coordinate pair that is the solution, where both x and y are the same. Normally, when you have a system and are solving them simultaneously, you are looking for the point at which they are equal. This is a very useful concept in business and finance, both in the home for personal information, and in the office setting where companies are. Where the 2 equations intersect is a point where they are equal.
Part B:
The graphs do not intersect right at a perfect integer of x. Therefore, we will solve these equations simultaneously to solve first for x, then we will plug in x to solve for y. Since we have the equations set to equal each other, we can solve for x by getting everything on one side of the equation and then setting it equal to 0.
2 - x = 4x + 3 so
5x + 1 = 0. Solving for x,
5x = -1 so
The y coordinate can be found by subbing in this value of x into either equation. If y = 2 - x, and x = -1/5, then
y = 2 -(-1/5) and y = 2 + 1/5 and y = 10/5 + 1/5 gives us that y = 11/5
Thus, the coordinate pair that is the solution to that system is
Part C:
You would solve the system graphically by graphing both lines on the same window. However, since their intersection is not an integer pair, but are fractions, you would not be able to tell EXACTLY where they intersect. From the graphing window, you would hit your 2nd button then "trace" which is in the row at the very top of the buttons below the window. Then hit 5: intersect. You'll be back to your graph of the lines, and there will be a cursor blinking along the line you graphed under Y1. Move the cursor til it is right over the intersection of the lines and hit "enter". Then you'll be back to the graphs with a blinking cursor over the line you entered in Y2. Move that cursor along the line til it is dead-center over the other point on the first line and hit "enter" again. At the bottom, you will see the x and y coordinates that are the intersection of this system.
You might be interested in
Answer:
(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.
(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.
(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.
(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.
Step-by-step explanation:
To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.
The significance level for the test is <em>α</em> = 0.05.
The decision rule is:
If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.
(a)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>
The sample size is, <em>n</em> = 11.
The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.
The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₁₀ > 1.91) = 0.043.
The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.
The null hypothesis is rejected at 5% level of significance.
(b)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>
The sample size is, <em>n</em> = 17.
The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.
The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.
The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.
The null hypothesis is rejected at 5% level of significance.
(c)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>
The sample size is, <em>n</em> = 7.
The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.
The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.
The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.
The null hypothesis is not rejected at 5% level of significance.
(d)
The alternate hypothesis is:
<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>
The sample size is, <em>n</em> = 28.
The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.
The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.
Use a <em>t</em>-table t compute the <em>p</em>-value.
For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:
The <em>p</em>-value is:
P (t₂₇ > 2.12) = 0.022.
The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.
The null hypothesis is rejected at 5% level of significance.
Answer:
Step-by-step explanation:
we have
we know that
In the function
The variable of the function f is now the function g(x)
substitute
Answer: (0.8468, 0.8764)
Step-by-step explanation:
Formula to find the confidence interval for population proportion is given by :-
, where = sample proportion.
z* = Critical value
n= Sample size.
Let p be the true proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.
Given : Sample size = 3597
Number of students believe that they will find a job immediately after graduation= 3099
Then,
We know that , Critical value for 99% confidence interval = z*=2.576 (By z-table)
The 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation will be
Hence, the 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation. = (0.8468, 0.8764)