• ### Could anyone help me solve this probability problem?

Best answer: There are 38 20¢ coins, x 10¢ coins, (2x - 15) 50¢ coins. So there is a total of: 38 + x + (2x - 15) coins Simplifying this, we get: 38 + x + 2x - 15 3x + 23 total coins. The probability of pulling a 50¢ coin at random is the number of 50 cent coins (2x - 15) over the total number of coins (3x + 23), so: (2x - 15) / (3x + 23) And... show more
Best answer: There are 38 20¢ coins, x 10¢ coins, (2x - 15) 50¢ coins. So there is a total of: 38 + x + (2x - 15) coins Simplifying this, we get: 38 + x + 2x - 15 3x + 23 total coins. The probability of pulling a 50¢ coin at random is the number of 50 cent coins (2x - 15) over the total number of coins (3x + 23), so: (2x - 15) / (3x + 23) And we are told this is 1/8. So knowing this we can set the expression above equal to this and solve for x: (2x - 15) / (3x + 23) = 1/8 Multiply both sides by the LCD of 8(3x + 23): 8(2x - 15) = 3x + 23 16x - 120 = 3x + 23 13x = 143 x = 11 So now we know what x is, therefore there are: 11 10¢ coins and (2 * 11 - 15) (or 7) 50¢ coins. So the total number of coins is: 38 + 11 + 7 = 56 So the probability of drawing a 20¢ coin is: 38/56 = 19/28 And the probability of drawing a 10¢ coin is: 11/56 If we did everything right, if we add up these probabilities, we should get a total of 1 since we've accounted for every possible outcome: 1/8 + 19/28 + 11/56 LCD: 7/56 + 38/56 + 11/56 add the numerators: 56/56 1 And it works out, so again your answers are: x = 11 The probability of drawing a 20¢ coin is 19/28 The probability of drawing a 10¢ coin is 11/56
1 answer · Mathematics · 10 hours ago
• ### Maths Question (Triangle)?

Best answer: The lengths of all three legs are not the same. Two are 2√2 (if you do the math you will see that to be true) and the other is 4 units. So it is not an equilateral triangle. Since two of the sides are the same size, it is Isosceles. Scalene is the term used if all their sides are a different length. So no triangle is ever more than... show more
Best answer: The lengths of all three legs are not the same. Two are 2√2 (if you do the math you will see that to be true) and the other is 4 units. So it is not an equilateral triangle. Since two of the sides are the same size, it is Isosceles. Scalene is the term used if all their sides are a different length. So no triangle is ever more than one of options A, B, or C. As for the last one, we can test it with Pythagorean Theorem to see if it's a right triangle or not. If both sides are equal, then it is. I said the lengths of the smaller two sides were 2√2 and the larger side is 4, so: a² + b² = c² (2√2)² + (2√2)² = 4² 4 * 2 + 4 * 2 = 16 8 + 8 = 16 16 = 16 TRUE So this is a right triangle. So your answers are options B and D. ------ In case you don't know how I knew the smaller sides were 2√2, if we take a point on the base and the point on top and look at that as its own right triangle, we can see the bottom side has a length of 2 and the right side has a length of 2, so we can solve for the hypotenuse to get the length of the smaller side of the original traingle: a² + b² = c² 2² + 2² = c² 4 + 4 = c² c² = 8 c = √8 c = 2√2
3 answers · Mathematics · 4 months ago
• ### Solving non-factorable inequalities? x^-5>0 and 3x^3+8x^2-x<2?

Best answer: You still solve them the same way. You factor them to find the roots to use as pivots. If they aren't factorable, you still find the roots and use them as pivots, but your roots aren't rational: x² - 5 > 0 We can add 5 to both sides: x² > 5 Square root of both sides. In an equation, you would put a ± on the other... show more
Best answer: You still solve them the same way. You factor them to find the roots to use as pivots. If they aren't factorable, you still find the roots and use them as pivots, but your roots aren't rational: x² - 5 > 0 We can add 5 to both sides: x² > 5 Square root of both sides. In an equation, you would put a ± on the other side. Since this is an inequality, the - side has the sign invert: x > 5 and x < -5 So you have all real numbers except for -5, 5, and all values in between. ----------------- 3x³ + 2x² - x < 2 I started out with subtracting 2 from both sides: 3x³ + 2x² - x - 2 < 0 Then I graphed it and found there to be one irrational root, but that irrational number isn't an easy number to represent. So your best bet would be to use something like Newton's method to determine what that root is. Since you are looking for values under zero (negative numbers), then whatever that root is, based on the graph, will be any values of x less than that value. It ends up being about: x < 0.79799
4 answers · Mathematics · 2 days ago

Best answer: Is it illegal? Yes. Is anyone really going to press charges on you? Probably not. These are games that are 13-15 years old that are not getting sold as new today, so Nintendo isn't "losing money" on this. But yeah, it's not legal. If it makes you feel better, find an old GameCube or GC-compatible Wii and track... show more
Best answer: Is it illegal? Yes. Is anyone really going to press charges on you? Probably not. These are games that are 13-15 years old that are not getting sold as new today, so Nintendo isn't "losing money" on this. But yeah, it's not legal. If it makes you feel better, find an old GameCube or GC-compatible Wii and track down working discs and play them on your TV as they were intended.
3 answers · Video & Online Games · 2 days ago
• ### My Nintendo Wii sensor won’t Sense my cursor.?

Best answer: The sensor bar is really just an IR emitter and the cord that plugs into the Wii is just to supply power. So sounds like the sensor bar port in the new Wii isn't working and not delivering power to the sensor bar. If you still have the old Wii, you can have that running next to the new one, turned on, to supply power to the sensor... show more
Best answer: The sensor bar is really just an IR emitter and the cord that plugs into the Wii is just to supply power. So sounds like the sensor bar port in the new Wii isn't working and not delivering power to the sensor bar. If you still have the old Wii, you can have that running next to the new one, turned on, to supply power to the sensor bar and that will work just fine. Or find one of the alternate sensor bar solutions that are out there (USB-powered that you plug into a port on your TV or the Wii, or a wireless one that is powered by batteries).
1 answer · Nintendo Wii · 3 days ago
• ### So I've been wanting to upgrade to an xbox one...?

Best answer: Not all games are backward compatible, but a lot of games are. Those that are, if you digitally bought them for 360, the can just be downloaded on the XBO and will work. If you have the disc, putting the disc into the system will start the process of downloading the XBO version so you can play it. The list of games that this works on... show more
Best answer: Not all games are backward compatible, but a lot of games are. Those that are, if you digitally bought them for 360, the can just be downloaded on the XBO and will work. If you have the disc, putting the disc into the system will start the process of downloading the XBO version so you can play it. The list of games that this works on can be found here: https://www.xbox.com/en-US/xbox-one/back... There are 515 games on the list right now and they add to it often.
1 answer · Xbox · 3 days ago
• ### Nintendo e-shop game recommendations?

Best answer: If you like Puzzle games, I put in over 160 hours into Pic-a-Pix Deluxe. It's Picross with colors that comes with 150 puzzles with tons of DLC content ranging from \$3 - \$5 each if you want more, so you can buy them as you want. If you like older arcade games, there is Arcade Archives: Donkey Kong, which is the first time an... show more
Best answer: If you like Puzzle games, I put in over 160 hours into Pic-a-Pix Deluxe. It's Picross with colors that comes with 150 puzzles with tons of DLC content ranging from \$3 - \$5 each if you want more, so you can buy them as you want. If you like older arcade games, there is Arcade Archives: Donkey Kong, which is the first time an arcade-accurate port of the game has been released anywhere. They also have an Arcade Archives release of Mario Bros. For indie-action games, Stick It to the Man is a good one. Golf Story is reminiscent of the old Mario Golf RPG games from GameBoy era with good humor and the best use of the rumble feature of the Joy-Con so far. :) And the last of the games that I'll recommend, if you like party games, there are four games in the Jack Box Party Pack series out with a fifth one coming. If you were to only choose one, go with the 3rd pack first as the Trivia Murder Party is probably the best single game in the franchise. Playable by up to 8 people (most games, some are 4 or 6) everyone uses their own phones or other web-enabled devices as their controllers (so you don't need to invest in 8 Joy-Con to play). If you like 3 and want more, go with 4 next. 1 and 2 have games that are pretty much obsoleted with 3 and 4 with better and more current versions with the exception of the original You Don't Know Jack game, which will get an updated version with Pack 5.
1 answer · Video & Online Games · 3 days ago
• ### Can i run Sims 4 with only 1.5 gigabytes of RAM?

Best answer: It says the minimum RAM is 2 GB. And the minimum integrated video card it supports has over 300 MB of RAM, so you may be shy on that front, too. Time to upgrade that old PC of yours.
Best answer: It says the minimum RAM is 2 GB. And the minimum integrated video card it supports has over 300 MB of RAM, so you may be shy on that front, too. Time to upgrade that old PC of yours.
2 answers · Video & Online Games · 5 days ago
• ### Im lagging very bad on fortnite why?

Best answer: You've got an old non-gaming CPU. i3 is the low-end of processors, and it's 3rd gen, so it's 5 years old. You don't say what your video card is, but if it's got a similar quality (meaning, lack-there-of), then it probably doesn't meet the minimum specs of the game.
Best answer: You've got an old non-gaming CPU. i3 is the low-end of processors, and it's 3rd gen, so it's 5 years old. You don't say what your video card is, but if it's got a similar quality (meaning, lack-there-of), then it probably doesn't meet the minimum specs of the game.
3 answers · PlayStation · 1 week ago
• ### Maths Help - Surface Area?

Best answer: If we look at the cross-section of this roof using the ceiling as the base, with a height of 2.5m from that to the top, we get an isosceles triangle with a base of 6m (using the 8m distance as a depth, so we can ignore that for now, but we'll get back to that). If we cut that in half down the top we have a right triangle with a base... show more
Best answer: If we look at the cross-section of this roof using the ceiling as the base, with a height of 2.5m from that to the top, we get an isosceles triangle with a base of 6m (using the 8m distance as a depth, so we can ignore that for now, but we'll get back to that). If we cut that in half down the top we have a right triangle with a base of 3m and the unchanged height of 2.5m Now we can find the length of the hypotenuse, or the slant height, which is what we need to find the surface area. a² + b² = c² 3² + 2.5² = c² 9 + 6.25 = c² 15.25 = c² c ≈ 3.905125m (rounding to a few extra DP to limit rounding errors). So now that we take the 8m depth in mind, we have two rectangles with dimensions of 3.905m by 8m. We can find the area of each, multiply by 2 to get the total area, then multiply it by the cost per m² to get the total cost of this roof: 3.905125 * 8 = 31.241 m² 31.241 * 2 = 62.482 m² 62.482 m² (32.5 £/m²) = £2,030.67 (rounded to the nearest pence)
1 answer · Mathematics · 6 days ago
• ### Vectors Proof?

Best answer: When multiplying two numbers together, it's the same as adding one value to itself the second value's times, example: 2 * 3 = 2 + 2 + 2 and 3 + 3 So if we have: a * (b + c) We can add (b + c) to itself "a" times: (b + c) + (b + c) + (b + c) + ... + (b + c) Since that is done "a" times, if we then... show more
Best answer: When multiplying two numbers together, it's the same as adding one value to itself the second value's times, example: 2 * 3 = 2 + 2 + 2 and 3 + 3 So if we have: a * (b + c) We can add (b + c) to itself "a" times: (b + c) + (b + c) + (b + c) + ... + (b + c) Since that is done "a" times, if we then simplify by combining like terms, we get: ab + ac so: a(b + c) = ab + ac
2 answers · Mathematics · 1 week ago
• ### I NEED MATHS HELP RIGHT NOW IM SO STUCK IM JUST NOT ABLE TO EVEN UNDERSTAND IT PLEASE HELP ITS DUE TOMORROW!!!?

Best answer: Prove that x(x - 1) > 6(x - 2) for all x ≥ 5 Let's start with that inequality and simplify it to see what happens: x(x - 1) > 6(x - 2) x² - x > 6x - 12 x² - 7x + 12 > 0 This factors: (x - 3)(x - 4) > 0 So we can use the roots of 3 and 4 as pivots to find ranges of x that results in the product to be greater than 0... show more
Best answer: Prove that x(x - 1) > 6(x - 2) for all x ≥ 5 Let's start with that inequality and simplify it to see what happens: x(x - 1) > 6(x - 2) x² - x > 6x - 12 x² - 7x + 12 > 0 This factors: (x - 3)(x - 4) > 0 So we can use the roots of 3 and 4 as pivots to find ranges of x that results in the product to be greater than 0 (positive values): if x > 4 we have positive times positive which is positive, so in our solution set. if 3 < x < 4 we have positive times negative which is negative, so not in our solution set. if x < 3 we have negative times negative which is positive, so in our solution set. So the full solution of this inequality is: x < 3 and x > 4 You are asked to show that if x ≥ 5, this is true. Since 5 > 4, and any number greater than 5 is also greater than 4, then this shows that x ≥ 5 is also true as a subset of the full solution.
2 answers · Mathematics · 1 week ago
• ### Does wifi latency affect PS4 remote play on PC?

Best answer: I've been using Remote Play a lot lately, but since I have a 1 ms ping to my provider, even if it does go out and back in I wouldn't notice it. If the PS4 and the PC software are smart enough to know that they are on the same network, the it won't go out through the satellite and come back. You won't have problems with... show more
Best answer: I've been using Remote Play a lot lately, but since I have a 1 ms ping to my provider, even if it does go out and back in I wouldn't notice it. If the PS4 and the PC software are smart enough to know that they are on the same network, the it won't go out through the satellite and come back. You won't have problems with it if this is the case as it's just a local connection at that point. If the network signal does go out into the cloud and come back in, then yes, you will have issues . 700ms means that if you are looking at something that is almost a second old. If you react to it and press a button, it won't be received by your system for another "almost a second". So you will be playing a game 1.5 seconds behind. If it was a turn-based fighting game, you may learn to live with it. But if it's got any sort of twitch combat, it won't work.
1 answer · Video & Online Games · 1 year ago
• ### Are there any good pokemon games on Xbox 360?

Best answer: Pokemon has only ever been and will only ever be a game series released on Nintendo platforms. Nintendo is a partial owner of The Pokemon Company. There are many Pokemon games available for the 3DS family of systems. The cheapest way to get started would be to look for a 2DS (not the "new 2DS XL") which should cost about \$70... show more
Best answer: Pokemon has only ever been and will only ever be a game series released on Nintendo platforms. Nintendo is a partial owner of The Pokemon Company. There are many Pokemon games available for the 3DS family of systems. The cheapest way to get started would be to look for a 2DS (not the "new 2DS XL") which should cost about \$70 - \$80 for the system and usually comes with a game pre-loaded (look for a system that looks like a door stop wedge vs. a clam-shell system. Then the Pokemon games usually cost about \$40. There is a new Pokemon game coming out this fall for the Nintendo Switch, but that is a much larger cost (\$300 for the system and \$60 for the game) and not really sure that I would get a Switch for one so young.
4 answers · Xbox · 1 week ago
• ### Why does microsoft charge for everything on Xbox?!?

Best answer: Charging for a name change discourages people from changing their name willy-nilly. If they could, someone would change their name to something silly, grief players, then get the name changed again to escape punishment.
Best answer: Charging for a name change discourages people from changing their name willy-nilly. If they could, someone would change their name to something silly, grief players, then get the name changed again to escape punishment.
3 answers · Corporations · 1 week ago
• ### Has Kelly or anyone already played the game?

Best answer: Who's Kelly and what game? This isn't a personal message board. It's a very public (worldwide) question/answer board.
Best answer: Who's Kelly and what game? This isn't a personal message board. It's a very public (worldwide) question/answer board.
1 answer · Video & Online Games · 2 weeks ago
• ### The quotient of -18x^6/6x^3 is equal to A-3x^3 B-3x^2 C-12x^2 D-12x^3?

Best answer: As written: -18x⁶ / 6x³ Only the 6 is divided by the numerator, then the x³ is multiplied by the result, so the result would be: -3x⁹ So presuming you mean: -18x⁶ / (6x³) then the x³ gets divided by the numerator to get: -3x³ (answer A)
Best answer: As written: -18x⁶ / 6x³ Only the 6 is divided by the numerator, then the x³ is multiplied by the result, so the result would be: -3x⁹ So presuming you mean: -18x⁶ / (6x³) then the x³ gets divided by the numerator to get: -3x³ (answer A)
4 answers · Mathematics · 2 weeks ago
• ### If I use multiple Xbox gift cards, will they all add up?

Best answer: I don't know what the limit is for XBL, but for PSN and eShop, the limit is \$200. You can stack as many cards on top of each other as long as you don't go past the upper limit.
Best answer: I don't know what the limit is for XBL, but for PSN and eShop, the limit is \$200. You can stack as many cards on top of each other as long as you don't go past the upper limit.
2 answers · Xbox · 2 weeks ago
• ### Calculus 2 question, Newton's Cooling?

Best answer: Let's see what I get. I use this equation: T(t) = Tₐ + (T₀ - Tₐ)e^(-kt) Where: T(t) is the temp after "t" "time units" (can be seconds, minutes, hours, days, etc. depending on what you are doing) Tₐ is the ambient temperature of the environment the object is placed in T₀ is the temperature of the object at t =... show more
Best answer: Let's see what I get. I use this equation: T(t) = Tₐ + (T₀ - Tₐ)e^(-kt) Where: T(t) is the temp after "t" "time units" (can be seconds, minutes, hours, days, etc. depending on what you are doing) Tₐ is the ambient temperature of the environment the object is placed in T₀ is the temperature of the object at t = 0 k is the constant (usually have to solve for first) t is the time in whatever "time units" is required. So let's see what we have: A hot metal bar is submerged in a large reservoir of water with a temperature of 70℉ (Tₐ). The temperature after 20 s after submersion is 100℉ (T(20)). After 1 minute submerged, the temperature has cooled to 80℉ (T(60)) So we have two data points with an unknown k and an unknown T₀. So we set up a system of two equations. So we have: T(t) = Tₐ + (T₀ - Tₐ)e^(-kt) T(20) = 70 + (T₀ - 70)e^(-k * 20) and T(60) = 70 + (T₀ - 70)e^(-k * 60) 100 = 70 + (T₀ - 70)e^(-20k) and 80 = 70 + (T₀ - 70)e^(-60k) 30 = (T₀ - 70)e^(-20k) and 10 = (T₀ - 70)e^(-60k) To make things easy, I'll drop the subscripts now that we have two variables. So to solve for k, we want to solve one equation for T in terms of k, substitute into the other equation, and solve for k: 30 = (T - 70)e^(-20k) 30 / e^(-20k) = T - 70 70 + 30 / e^(-20k) = T Substituting into the other equation: 10 = (T - 70)e^(-60k) 10 = (70 + 30 / e^(-20k) - 70)e^(-60k) 10 / e^(-60k) = 70 + 30 / e^(-20k) - 70 10 / e^(-60k) = 30 / e^(-20k) reciprocal of both sides, then multiply both sides by 30: e^(-60k) / 10 = e^(-20k) / 30 3e^(-60k) = e^(-20k) natural log of both sides: ln[3e^(-60k)] = -20k The log of a product is the same as the sum of the logs: ln(3) + ln[e^(-60k)] = -20k ln(3) + (-60k) = -20k ln(3) = 40k k = ln(3) / 40 So I got what you got for part 1. Now let's use this to solve for T (your part d): 10 = (T - 70)e^(-60k) 10 = (T - 70)e^(-60 * ln(3) / 40) 10 = (T - 70)e^(-3 ln(3) / 2) 10 / (T - 70) = e^(-3 ln(3) / 2) Natural log of both sides: ln[10 / (T - 70)] = -3 ln(3) / 2 Log of the quotient is the same as the difference of two logs: ln(10) - ln(T - 70) = -3 ln(3) / 2 Let's move the constant log to the right side, then multiply both sides by -1: -ln(T - 70) = -3 ln(3) / 2 - ln(10) ln(T - 70) = 3 ln(3) / 2 + ln(10) I normally don't like to use approximations until the end, but at this point, I'll get a decimal approximation for the right side to about 7 DP: ln(T - 70) = 3.9505035 Now have both sides be an exponent over a base of "e" to cancel out the log: T - 70 = e³·⁹⁵⁰⁵⁰³⁵ Turn that into a decimal approximation to 7 DP, then add 70 to both sides: T - 70 = 51.9615229 T = 121.9615229 So for two DP, I get 121.96℉. So you may have rounding errors if you did mid-step rounding before the end.
3 answers · Mathematics · 3 weeks ago
• ### The wind chill temperature when the outside temperature in 20F is given by?

Best answer: You are given the function: f(x) = 0.0053x² - 0.65x + 15.3 To determine the wind chill when the temp is 20℉ and x is the wind speed in miles per hour. You want to know what wind speed will make the windchill less than or equal to -2℉. So we can set up an inequality: f(x) ≤ -2 Substitute the function to get: 0.0053x² - 0.65x + 15.3... show more
Best answer: You are given the function: f(x) = 0.0053x² - 0.65x + 15.3 To determine the wind chill when the temp is 20℉ and x is the wind speed in miles per hour. You want to know what wind speed will make the windchill less than or equal to -2℉. So we can set up an inequality: f(x) ≤ -2 Substitute the function to get: 0.0053x² - 0.65x + 15.3 ≤ -2 Now let's add 2 to both sides so we have a zero on the right: 0.0053x² - 0.65x + 17.3 ≤ 0 We need to find the roots of this to "pseudo-factor" the expression. So using quadratic equation we get: x = [ -b ± √(b² - 4ac)] / (2a) x = [ -(-0.65) ± √((-0.65)² - 4 * 0.0053 * 17.3)] / (2 * 0.0053) x = [ 0.65 ± √(0.4225 - 0.36676)] / 0.0106 x = (0.65 ± √0.05574) / 0.0106 x = (0.65 ± 0.23609) / 0.0106 x = (0.65 - 0.23609) / 0.0106 and (0.65 + 0.23609) / 0.0106 x = 0.41391 / 0.0106 and 0.88609 / 0.0106 x = 39.05 and 83.59 (rounded to 2DP) So if these are the roots, then the factors would be: (x - 39.05)(x - 83.59) ≤ 0 Now we can use these as pivots to look at ranges to see what values gives us a number less than 0 (negative numbers) When x > 83.59, we have positive times positive, which is positive, so not in the solution set. When 39.05 < x < 83.59, we have positive times negative, which is negative, in our solution set. when x < 39.05 we have negative times negative, which is positive, so not in the solution set. And since we're looking for less than OR EQUAL TO, the values at the roots are also in the solution set. So the wind chill is -2℉ or below when the wind is between 39.05 mph and 83.59 mph. While this isn't completely true in how wind chill really works (it's more logarithmic vs. parabola as it won't get warmer if the wind is faster, but also doesn't have a measurable effect when more than 45 mph), for this exercise, this is the answer you are looking for.
2 answers · Mathematics · 3 weeks ago