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3 years ago

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A bag contains 3 blue marbles, 1 green marble, and 4 orange marbles. A marble is pulled out of the bag at random. The probabilit

y that the marble is blue is____.
A. 3/5
B. 0
C. 5/8
D. 3/8
E. 3/4

1 answer:

Ainat [17]3 years ago

7 0

Answer:

3\49857

Step-by-step explanation:

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Solve the following matrix equations: (matrices)

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Step-by-step explanation:

a)

$3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 & 6 \\ 10 & 14 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} \\ \\ 3X = \begin{pmatrix} - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\ \\ 3X = \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \frac{1}{3} \begin{pmatrix} - 3 & 3 \\ 6 & 9\end{pmatrix}\\ \\ X = \begin{pmatrix} \frac{ - 3}{3} & \frac{3}{3} \\ \\ \frac{6}{3} & \frac{9}{3} \end{pmatrix}\\ \\ \huge \red{ X} = \purple{ \begin{pmatrix} - 1 &1 \\ 2 & 3 \end{pmatrix}}$

b)

$3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X =\begin{pmatrix} \frac{3}{3} & \frac{0}{3} & \frac{-3}{3} \\\\ \frac{6}{3} & \frac{3}{3} & \frac{0}{3} \\\\ \frac{9}{3} & \frac{6}{3} & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1 & 0 & - 1\\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}}\\$

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The equation 7^2=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

jeyben [28]

Answer:

The period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

Step-by-step explanation:

The equation $T ^ 2 = a ^ 3$ shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?

For the planet Y:

$T_y ^ 2 = a_y ^ 3$

For planet X:

$T_x ^ 2 = a_x ^ 3$

To know the factor of aumeto we compared $T_x$ with $T_y$

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:

$a_y ^ 3 = (a_xk) ^ 3$

So

$\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x$

Then, the period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

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