George desposited 2000 in his bank, which offers 5% compoud interest annually. What would be the principla amount available in g
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I made my own graph using the given points. Pls. see attachment.
<span>Part A: Using the graph above, create a system of inequalities that only contains points D and F in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above.
I used paint program to graph the points based on the given data.
I used a ruler as guide as I draw my line containing both points D and F and extended the line outward.
The yellow lines above represents the shaded portion of the graph.
y = mx + b
b is the y-intercept. It is the value of y when x is 0. In this case, b = 3.5
m is the slope. It is the change in y divided by the change in x.
point D (2,4) ; point F (-2,3)
change in y: 4 - 3 = 1
change in x: 2 - (-2) = 4
m = 1/4
y = 1/4 x + 3.5
Since we are looking for a system of inequality and the portion shaded is in the upper part of the line, the value of y will be greater than or equal to.
y </span>≥ 1/4x + 3.5<span>
Part B: Explain how to verify that the points D and F are solutions to the system of inequalities created in Part A.
To check the solution given above. We will use the x values of points D and F and check whether its y values are equal to or greater than the solution.</span>
y ≥ 1/4x + 3.5
Point D (2,4) : x = 2 : y ≥ 1/4 * 2 + 3.5 → 0.5 + 3.5 → 4
Point F (-2.3): x = -2 ; y ≥ 1/4 * -2 + 3.5 → -0.5 + 3.5 → 3
<span>
Part C: Chickens can only be raised in the area defined by y > 2x − 2. Explain how you can identify farms in which chickens can be raised.
y > 2x - 2
We will get the coordinates of each point and see if its x value will generate the desired value of y.
A(2,-3) ; x = 2 : </span><span>y > 2(2) - 2 </span>→ 4 - 2 → 2 . DID NOT PASS. -3 is lesser than 2.<span>
B(-3,-4) ; x = -3 : y > 2(-3) - 2 </span>→ -6 - 2 → -8. PASSED. -4 is greater than -8.<span>
C(-4,2) ; x = -4 : y > 2(-4) - 2 </span>→ -8 - 2 → -10. PASSED. 2 is greater than -10.<span>
D(2,4) ; x = 2 : y > 2(2) - 2 </span>→ 4 - 2 → 2. PASSED. 4 is greater than 2.<span>
E(3,1) ; x = 3 : y > 2(3) - 2 </span>→ 6 - 2 → 4. DID NOT PASS. 1 is lesser than 4.<span>
F(-2,3) ; x = -2 ; y > 2(-2) - 2 </span>→ -4 - 2 → -6. PASSED. 3 is greater than -6.
Among the farms, points B,C, D, and F can raise chickens based on the given system of linear inequality.
<span>Part A: Using the graph above, create a system of inequalities that only contains points D and F in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above.
I used paint program to graph the points based on the given data.
I used a ruler as guide as I draw my line containing both points D and F and extended the line outward.
The yellow lines above represents the shaded portion of the graph.
y = mx + b
b is the y-intercept. It is the value of y when x is 0. In this case, b = 3.5
m is the slope. It is the change in y divided by the change in x.
point D (2,4) ; point F (-2,3)
change in y: 4 - 3 = 1
change in x: 2 - (-2) = 4
m = 1/4
y = 1/4 x + 3.5
Since we are looking for a system of inequality and the portion shaded is in the upper part of the line, the value of y will be greater than or equal to.
y </span>≥ 1/4x + 3.5<span>
Part B: Explain how to verify that the points D and F are solutions to the system of inequalities created in Part A.
To check the solution given above. We will use the x values of points D and F and check whether its y values are equal to or greater than the solution.</span>
y ≥ 1/4x + 3.5
Point D (2,4) : x = 2 : y ≥ 1/4 * 2 + 3.5 → 0.5 + 3.5 → 4
Point F (-2.3): x = -2 ; y ≥ 1/4 * -2 + 3.5 → -0.5 + 3.5 → 3
<span>
Part C: Chickens can only be raised in the area defined by y > 2x − 2. Explain how you can identify farms in which chickens can be raised.
y > 2x - 2
We will get the coordinates of each point and see if its x value will generate the desired value of y.
A(2,-3) ; x = 2 : </span><span>y > 2(2) - 2 </span>→ 4 - 2 → 2 . DID NOT PASS. -3 is lesser than 2.<span>
B(-3,-4) ; x = -3 : y > 2(-3) - 2 </span>→ -6 - 2 → -8. PASSED. -4 is greater than -8.<span>
C(-4,2) ; x = -4 : y > 2(-4) - 2 </span>→ -8 - 2 → -10. PASSED. 2 is greater than -10.<span>
D(2,4) ; x = 2 : y > 2(2) - 2 </span>→ 4 - 2 → 2. PASSED. 4 is greater than 2.<span>
E(3,1) ; x = 3 : y > 2(3) - 2 </span>→ 6 - 2 → 4. DID NOT PASS. 1 is lesser than 4.<span>
F(-2,3) ; x = -2 ; y > 2(-2) - 2 </span>→ -4 - 2 → -6. PASSED. 3 is greater than -6.
Among the farms, points B,C, D, and F can raise chickens based on the given system of linear inequality.
Answer:
The critical value can be founded with the following Excel formula:
=CHISQ.INV(1-0.05,5)
And we got
a. 11.070
And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Solution to the problem
For this case we want to test:
H0: Absenteeism is distributed evenly throughout the week
H1: Absenteeism is NOT distributed evenly throughout the week
We have the following data:
Monday Tuesday Wednesday Thursday Friday Saturday Total
12 9 11 10 9 9 60
The level of significance assumed for this case is
The statistic to check the hypothesis is given by:
The table given represent the observed values, we just need to calculate the expected values with the following formula and the expected value is the same for all the days since that's what we want to test.
now we can calculate the statistic:
Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:
The critical value can be founded with the following Excel formula:
=CHISQ.INV(1-0.05,5)
And we got
a. 11.070
And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance