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3 years ago

A card is chosen at random from a deck of 52 cards. What is the probability that a king or queen is chosen?

Mathematics

1 answer:

Slav-nsk [51]3 years ago

4 0

The answer is B) 2/13.

An easy way to figure this out is to find what the unsimplified probability would be:

Since there are 8 kings and queens in total, the number is:

8/52

But, since these are both even numbers, they can be simplified.

1/4 x 8 = 2

1/4 x 52 = 13

2/13

Since 13 is a prime number, the answer cannot be simplified anymore.

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Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

otez555 [7]

Answer:

$D = \frac{25\sqrt{13}}{13}$

Step-by-step explanation:

Given

$y = 4 - \frac{2}{3}x$

$(x_1,y_1) = (-2,-3)$

Required

Determine the distance

$y = 4 - \frac{2}{3}x$

Write the above equation in standard form:

$Ax + By + C = 0$

So, we have:

$\frac{2}{3}x+y - 4 = 0$

By comparison:

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

The distance is calculated using:

$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Where:

$(x_1,y_1) = (-2,-3)$

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

This gives:

$D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}$

$D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}$

Take LCM

$D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}$

$D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}$

$D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}$

$D = \frac{25}{3}/\sqrt{\frac{13}{9}}$

Split the square root

$D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}$

Change / to *

$D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}$

$D = \frac{25}{3}*\frac{3}{\sqrt{13}}$

$D = \frac{25}{\sqrt{13}}$

Rationalize

$D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}$

$D = \frac{25\sqrt{13}}{13}$

Hence, the distance is:

$D = \frac{25\sqrt{13}}{13}$

4 0

3 years ago