Use the expansion of (4x+3)^3 to find the exact value of (4.02)^3
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Answer:
We want to find a solution of the system:
x^2 + y^2 > 49
y ≤ –x^2 – 4
Here we do not have any options, so let's try to find a general solution.
First, we can remember that the equation of a circle centered in the point (a, b) and of radius R is:
(x - a)^2 + (y - b)^2 = R^2
If we look at our first inequality, we can write it as:
x^2 + y^2 > 7^2
So the solutions of the first inequality are all the points that are outside (because the symbol used is >) of the circle of radius R = 7 centered in the origin.
From the other equation, we would get:
y ≤ –x^2 – 4
This is parabola, anything that is in the graph of the parabola or below will be a solution for this inequality.
Then the solutions of the system, are the ones that are in the region of solutions for both inequalities.
You can see the graph below, where both regions are graphed. The intersection of these regions is the region of the solutions for the system of inequalities:
by looking at the graph, we can see a lot of points that are solutions, like:
(0, -10)
(0, -15)
(2, -10)
etc.
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