4p + 1 > −7 or 6p + 3 < 33? how do I solve this
1 answer:
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The graphs of f(x) and g(x) are transformed function from the function y = x^2
The set of inequalities do not have a solution
<h3>How to modify the graphs</h3>
From the graph, we have:
and
To derive y < x^2 - 3, we simply change the equality sign in the function f(x) to less than.
To derive y > x^2 + 2, we perform the following transformation on the function g(x)
- Shift the function g(x) down by 2 units
- Reflect across the x-axis
- Shift the function g(x) down by 3 units
- Change the equality sign in the function g(x) to greater than
The inequalities of the graphs become
y < x^2 - 3 and y > x^2 + 2
From the graph of the above inequalities (see attachment), we can see that the curves of the inequalities do not intersect.
Hence, the set of inequalities do not have a solution
Read more about inequalities at:
Answer:
Option (4)
Step-by-step explanation:
Vertices of the triangle are X(0, -4), Y(2, -3) and Z(2, -6).
Slope of XY () =
=
() =
Similarly, slope of XZ () =
() = (-1)
Which should be (-1) if XY and XZ are perpendicular to each other.
Now we can say that XY and XZ are not perpendicular.
Therefore, Lydia's assertion that ΔXYZ is a right triangle is not correct.
Option (4) will be the answer.
Answer:
c) Is not a property (hence (d) is not either)
Step-by-step explanation:
Remember that the chi square distribution with k degrees of freedom has this formula
Where N₁ , N₂m .... are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.
Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true
The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.