A tire with a radius of 22 centimeters

1 answer:

6 0

The answer will be 138.16

You will find the answer by multiplying by 2 and 22 which will give you 44, then you multiply that but 3.14 which will give you 138.16

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7. What type of lines does the<br> system y=3x+2 and 3x-y=-2<br> form?

Aleks [24]

Answer: The system forms the same line.

Step-by-step explanation:

3x - y = -2

-y = -3x - 2 (subtract 3x from both sides)

y = 3x + 2 (divide both sides by -1)

First equation: y = 3x +2

Second equation: y = 3x +2

THEY ARE THE SAME LINE —> there are infinitely many solutions.

3 0

3 years ago

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.)

Since the outcome of these two events are independent,

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

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:

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

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:

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