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Marat540 [252]
4 years ago
5

Find the equation of the surface. a standard cone with vertex at (6, 0 , 0) opening up on the positive x-axis.

Mathematics
1 answer:
DerKrebs [107]4 years ago
7 0

Solution: Since, it’s a standard cone opening up on the positive x-  axis  

Therefore, x=√y2+z2


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In quadrilateral $ABCD$, we have $AB=3,$ $BC=6,$ $CD=4,$ and $DA=4$.
vampirchik [111]
The triangle inequality applies.

In order for ACD to be a triangle, the length of AC must lie between CD-DA=0 and CD+DA=8.

In order for ABD to be a triangle, the length of AC must lie between BC-AB=3 and BC+AB=9.

The values common to both these restrictions are numbers between 3 and 8. Assuming we don't want the diagonal to be coincident with any sides, its integer length will be one of ...
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3 years ago
1. A wise old man once said, "500 reduced by 4 times my age is<br> 296." What is his age?
Daniel [21]

Answer:

51

Step-by-step explanation:

51×4=204

500-204=296

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M^3 is the square root of m6
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Two professors are applying for grants. Professor Jane has a probability of 0.61 of being funded. Professor Joe has probability
Anna11 [10]

Answer:

  • a. 0.1647;
  • b. 0.7153;
  • c. 0.4453.

Step-by-step explanation:

By definition, if two event A and B are independent, then

P(A\cap B) = P(A) \cdot P(B). (P(A\cap B) is the probability that the outcome of both event A and event B are true.)

<h3>a.</h3>

Since the outcome of these two events are independent,

\begin{aligned}P(\texttt{Jane} \cap \texttt{Joe}) &= P(\texttt{Jane}) \cdot P(\texttt{Joe})\\ &= 0.61 \times 0.27 \\&= 0.1647\end{aligned}.

<h3>b.</h3>

The logic not operator \lnot or the prime superscript ^{\prime} denotes that an event does not happen.  

P(\texttt{Jane}^{\prime}) = 1 - P(\texttt{Jane}) = 1- 0.61 = 0.39.

P(\texttt{Joe}^{\prime}) = 1 - P(\texttt{Joe}) = 1- 0.27 = 0.73.

Since the two events \texttt{Jane} and \texttt{Joe}, \texttt{Jane}^{\prime} and \texttt{Joe}^{\prime} are also independent. Probability that neither professor got funded:

P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = P(\texttt{Jane}^{\prime}) \cdot P(\texttt{Joe}^{\prime}) = 0.39 \times 0.73 = 0.2847.

Probability that at least one professor got funded- in other words, it is not true that neither professor got funded:

P((\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime})^{\prime}) = 1- P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = 0.7153.

<h3>c.</h3>

Similarly, since the two events \texttt{Jane} and \texttt{Joe}, \texttt{Jane} and \texttt{Joe}^{\prime} are also independent. Probability that Jane but not Joe got funded:

P(\texttt{Jane} \cap (\texttt{Joe}^{\prime})) = P(\texttt{Jane}) \cdot P(\texttt{Joe}^{\prime}) = 0.61 \times 0.73 = 0.4453.

4 0
3 years ago
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