there is a $10 monthly membership fee to download music. there is a $0.50 fee for each song download. what is the cost of downlo
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Answer:
Probability of picking an even and then even number =
Step-by-step explanation:
Probability of picking a card with even number,
P(even 1) = =
=
=
Followed by picking a card which is even without putting the first card back.
So an even card is to be picked out of remaining 3 cards.
P(even 2) =
=
Now probability of both events = P(even 1) × P(even 2)
=
=
Therefore, probability of both the events will be .
Check the picture below.
so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2} \\\\\\ AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AA%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20%0AC%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B8%7D%29%5Cqquad%20%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B%5B6-%28-4%29%5D%5E2%2B%5B8-%28-2%29%5D%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B%286%2B4%29%5E2%2B%288%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAC%3D%5Csqrt%7B10%5E2%2B10%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B10%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BAC%3D10%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2} \\\\\\ BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2} \\\\\\ BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0AB%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B6%7D%29%5Cqquad%20%0AD%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%5Cqquad%20%5Cqquad%20BD%3D%5Csqrt%7B%5B4-%28-2%29%5D%5E2%2B%5B0-6%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B%284%2B2%29%5E2%2B%28-6%29%5E2%7D%5Cimplies%20BD%3D%5Csqrt%7B6%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABD%3D%5Csqrt%7B6%5E2%282%29%7D%5Cimplies%20%5Cboxed%7BBD%3D6%5Csqrt%7B2%7D%7D)
that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.
namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,
![\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right] \\\\\\ 4[15]\implies 60](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20four%20congruent%20triangles%7D%7D%7B4%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%283%5Csqrt%7B2%7D%29%285%5Csqrt%7B2%7D%29%20%5Cright%5D%5Cimplies%204%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%2815%5Ccdot%20%28%5Csqrt%7B2%7D%29%5E2%29%20%5Cright%5D%7D%5Cimplies%204%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%2815%5Ccdot%202%29%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0A4%5B15%5D%5Cimplies%2060)
so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.
that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.
namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,