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kompoz [17]
3 years ago
12

a bag contains 9 marbles: 4 green 3 are red and 2 are blue what is the probability that the first marble is red and the second o

ne is blue write answer in simplest form.​
Mathematics
1 answer:
Korolek [52]3 years ago
3 0

Answer:

If the question states that they are not being replaced, then the answer is 1/8

If the question states that they are being replaced, then the answer is 1/9

Step-by-step explanation:

3/9 x 2/8 = 1/8

3/9 x 2/9 = 1/9

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8 students traveled in cars, which means that 450 students traveled by bus. 
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450 / 10 = 45
There were B. 45 students on each bus.
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If the following system of equations was written as a matrix equation in the form AX = C, and matrix A was expressed in the form
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5 0
3 years ago
2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?
7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of x^{2} which is 2.

The solutions to the quadratic equation 2x^2\:-\:5x\:+\:67\:=\:0 are:

x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

Step-by-step explanation:

Considering the equation

2x^2\:-\:5x\:+\:67\:=\:0

The first step is to divide all the terms by the coefficient of x^{2} which is 2.

so

\frac{2x^2-5x}{2}=\frac{-67}{2}

x^2-\frac{5x}{2}=-\frac{67}{2}

Lets now solve the equation by completeing the remaining steps

Write equation in the form: x^2+2ax+a^2=\left(x+a\right)^2

Solving for a,

2ax=-\frac{5}{2}x

a=-\frac{5}{4}

\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}

x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2

x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}

Completing the square

\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

For f^2\left(x\right)=a the solution are f\left(x\right)=\sqrt{a},\:-\sqrt{a}

Solving

x-\frac{5}{4}=\sqrt{-\frac{511}{16}}

x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}

x-\frac{5}{4}=i\sqrt{\frac{511}{16}}       ∵ Applying imaginary number rule \sqrt{-1}=i

x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}

-\frac{5}{4}=i\frac{\sqrt{511}}{4}

x=\frac{5}{4}+i\frac{\sqrt{511}}{4}

Similarly, solving

x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}

x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}    ∵ Applying imaginary number rule  \sqrt{-1}=i

x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

Therefore, the solutions to the quadratic equation are:

x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}

4 0
2 years ago
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