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3 years ago

Proof David Berkowitz is a psychopath 25 points + marked brainiest!

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

Answer:

According to the “Hare Psychopathy Checklist” a psychopath is a person who knows right from wrong in our society but often feels no guilt, remorse or empathy when they engage in wrong actions. They lie easily, blame others for their actions and often hurt people who are close to them. David Berkowitz meets the criteria of Hare’s checklist and is believed to be a true psychopath. Berkowitz is known as the “Son of Sam”, a thrill seeking serial killer who terrorized the city of New York for over a year. Information about David Berkowitz’s background, personality traits and lifestyle and his serial killer profile help to prove that he is in fact a psychopath. In conclusion, this essay will prove to readers that David Berkowitz is a notorious…

Step-by-step explanation:

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3 years ago

PLZZ HELPP!!

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There's one piece of information missing from the question, so
in order to completely calculate the answer, we must assume the
missing number.

We have no information that helps us calculate the probability
that you will eat the first jelly bean. So we'll have to assume that
you will definitely eat it, no matter what color it is. The probability is 1 .

-- You start out with (2 + 4 + 8 + 6) = 20 jelly beans in the package.
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-- If you close your eyes and pull one out of the package, then the
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-- Now there are 19 jelly beans left in the package.
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-- The probability of picking a blue one out of the package without peaking is

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3 years ago

Read 2 more answers

Which of the following is not a possible number of solutions when solving a system of equations containing a quadratic and a lin

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Answer:

Step-by-step explanation:

We have a system with two equations, one equation is a quadratic function and the other equation is a linear function.

To solve this system we have to clear "y" in both equations, and then equal both equations, then we will have a quadratic function and equal it to zero:

$ax^2+bx+c=0, a\neq 0$

Then to resolve a quadratic equation we apply Bhaskara's formula:

$x_{1}=\frac{-b+\sqrt{b^2-4ac} }{2a}$

$x_{2}=\frac{-b-\sqrt{b^2-4ac} }{2a}$

It usually has two solutions.

But it could happen that $\sqrt{b^2-4ac}$ then the equation doesn't have real solutions.

Or it could happen that there's only one solution, this happen when the linear equation touches the quadratic equation in one point.

And it's not possible to have more than 2 solutions. Then the answer ir 3.

For example:

In the three graphs the pink one is a quadratic function and the green one is a linear function.

In the first graph we can see that the linear function intersects the quadratic function in two points, then there are two solutions.

In the second graph we can see that the linear function intersects the quadratic function in only one point, then there is one solutions.

In the third graph we can see that the linear function doesn't intersect the quadratic function, then there aren't real solutions.

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Answer:

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