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Do 0844 numbers generate income?
In the UK, these numbers usually cost more than a normal phone call and can indeed generate income. Although they can cost up 40p/min to call, the revenue generated for the organisation being called will not be more than 5p/min, with the remainder retained by the telephone companies involved in connecting the call. Not all 0844 numbers pass on income to the business using them; sometimes the income is retained by the telephone company supplying the service. For example, a telephone company may often offer advanced call handling services, queuing, reporting, etc. to the business in return for the revenue they take from incoming calls. Subjectively, this could still be considered as generating income, as the company is receiving enhanced services it would otherwise have to pay the market rate for.
What is an 0871 number?
Calling an 0871 or 0872 number has three cost components: 1. The "service charge" or "revenue share" or "premium". This varies from 5p/min to 10p/min depending on exactly which number you call. Every caller, whether from landline or mobile, pays the same amount of "service charge". Ofcom maintains a list of these rates. 2. The "access charge" or "markup" or "profit". Callers from BT landlines pay ZERO for this component. BT is regulated to make no profit on calls to 0843, 0844, 0871 and 0872 numbers. Callers from other landlines pay 2p/min to 10p/min "access charge" and this varies depending on who you pay your phone bill to; e.g. Virgin is approx 4p/min. Callers using mobile phones pay 10p/min to 30p/min for this component, and it varies depending on which mobile operator you pay your phone bill to. 3. VAT VAT is currently 20% of the total price. Note: 0870 numbers currently work slightly differently. The revenue share for 0870 numbers was approx 5p/min but varied depending on the provider. Ofcom scrapped revenue sharing on 0870 numbers in 2009. The call price from landlines is usually about 5p/min to 10p/min but many landline providers allow these calls within inclusive minutes. Ofcom is likely to align 0870 pricing with that for 0871 and 0872 numbers in April or May 2013 with the introduction of "unbundled tariffs".
How many times do the hands of a clock overlap in a day?
A Better Approach (with reasoning)There are 2 cases depending on the working of the watch.Case 1: the movement of the second, minute and hour hands are continuous (not step-wise or click-based)Answer is: the hour and minute hands overlap every hour.Case 2 (very unusual): the hour hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the minute hand crosses (or reaches 12) and the minute hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the second hand crosses (or reaches 12).Answer: every 65 minutes. Interviewers often expect this answer cos they do not think accurately. The exact times are:0000 (12:00 AM)0105 (01:05 AM)0211 (02:11 AM)0316 (03:16 AM)0422 (04:22 AM)0527 (05:27 AM)0633 (06:33 AM)0738 (07:38 AM)0844 (08:44 AM)0949 (09:49 AM)1055 (10:55 AM)1200 (12:00 PM)1305 (01:05 PM)1411 (02:11 PM)1516 (03:16 PM)1622 (04:22 PM)1727 (05:27 PM)1833 (06:33 PM)1938 (07:38 PM)2044 (08:44 PM)2149 (09:49 PM)2255 (10:55 PM)Reasoning for Case 1:-----------------------------When do they overlap? At every (n + (n/11)) hours where n = 0, 1, 2, 3, ..., 24.How did I find this out? The following is the reasoning i used:At 0000, the hour and minute hands overlap. So number of overlaps now is 1. The minute hand races away and never again overlaps during the next one hour.Now, the minute hand moves at 360o/hour and the hour hand moves at 30o/hour.At 0100, the hour hand would be the 1 mark (or 30o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0100), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:30o + (30o)x(T) = (360o)x(T) How did I get to this equation? Note that, at 0100, when the minute hand starts moving from the 12 mark, the hour hand is already ahead of the minute hand by 30o. If the minute hand moves at a speed of 360o/hour, then in some time (T), it would cover (360o)(T) degrees. If the hour hand hand moves at a speed of 30o/hour, then in the same time (T), it would cover (30o)(T) degrees. Since the hour hand is already 30o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30o + the number of degrees covered in time T) which is (30o + (30o)x(T)).Now the condition when the two hands will overlap is that they should have covered the same number of degrees at a moment (or)No. of degrees covered by minute hand = No. of degrees covered by hour hand(or)30o + (30o)x(T) = (360o)x(T)If you solve this equation to find the value of T, you would get30o = (360o)x(T) - (30o)x(T)30o = (360o - 30o) x T30o = 330o x T(or)T = 30o/330oT = 1/11At 0200, the hour hand would be the 2 mark (or 60o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0200), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:60o + (30o)x(T) = (360o)x(T)Solving this equation, you will the value of T = 2/11At 0300, using the same reasoning (the hour hand at 90o past the 12 mark) and modifying the equation accordingly (90o + (30o)x(T) = (360o)x(T)), you would get the answer for T = 3/11.In general, the value for T for every hour is T = n/11 where n = 0, 1, 2, 3, ..., 24. So the exact time when the two hands overlap can be written as:the hour (n) + the time taken during that hour (T)(or)n + n/11where n = 0, 1, 2, 3, ..., 24.AM12:001:052:113:164:225:276:337:388:449:4910:55 PM12:001:052:113:164:225:276:337:388:449:4910:5522 is correct.The hands overlap about every 65 minutes, not every 60 minutes. In a day, the hands would only overlap 22 times, as illustrated in the table above.I would propose that the hands always overlap, as they're both attached at the center of the dial.If you didn't want to be facetious (or, at least, less facetious), you would still have to ask how many hands were on the clock. It may have a second hand, for example, or be digital (no hands at all).
How many times a day does a clock's hands overlap?
Answer is: the hour and minute hands overlap every 65 minutes. Interviewers often expect this answer because they do not think accurately. The exact times are:0000 (12:00 AM)0105 (01:05 AM)0211 (02:11 AM)0316 (03:16 AM)0422 (04:22 AM)0527 (05:27 AM)0633 (06:33 AM)0738 (07:38 AM)0844 (08:44 AM)0949 (09:49 AM)1055 (10:55 AM)1200 (12:00 PM)1305 (01:05 PM)1411 (02:11 PM)1516 (03:16 PM)1622 (04:22 PM)1727 (05:27 PM)1833 (06:33 PM)1938 (07:38 PM)2044 (08:44 PM)2149 (09:49 PM)2255 (10:55 PM)Case 1: the movement of the second, minute and hour hands are continuous (not step-wise or click-based)Case 2 (very unusual): the hour hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the minute hand crosses (or reaches 12) and the minute hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the second hand crosses (or reaches 12).Reasoning for Case 1:-----------------------------When do they overlap? At every (n + (n/11)) hours where n = 0, 1, 2, 3, ..., 24.How did I find this out? The following is the reasoning i used:At 0000, the hour and minute hands overlap. So number of overlaps now is 1. The minute hand races away and never again overlaps during the next one hour.Now, the minute hand moves at 360o/hour and the hour hand moves at 30o/hour.At 0100, the hour hand would be the 1 mark (or 30o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0100), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:30o + (30o)x(T) = (360o)x(T)How did I get to this equation? Note that, at 0100, when the minute hand starts moving from the 12 mark, the hour hand is already ahead of the minute hand by 30o. If the minute hand moves at a speed of 360o/hour, then in some time (T), it would cover (360o)(T) degrees. If the hour hand hand moves at a speed of 30o/hour, then in the same time (T), it would cover (30o)(T) degrees. Since the hour hand is already 30o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30o + the number of degrees covered in time T) which is (30o + (30o)x(T)).Now the condition when the two hands will overlap is that they should have covered the same number of degrees at a moment (or)No. of degrees covered by minute hand = No. of degrees covered by hour hand(or)30o + (30o)x(T) = (360o)x(T)If you solve this equation to find the value of T, you would get30o = (360o)x(T) - (30o)x(T)30o = (360o - 30o) x T30o = 330o x T(or)T = 30o/330oT = 1/11At 0200, the hour hand would be the 2 mark (or 60o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0200), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:60o + (30o)x(T) = (360o)x(T)Solving this equation, you will the value of T = 2/11At 0300, using the same reasoning (the hour hand at 90o past the 12 mark) and modifying the equation accordingly (90o + (30o)x(T) = (360o)x(T)), you would get the answer for T = 3/11.In general, the value for T for every hour is T = n/11 where n = 0, 1, 2, 3, ..., 24. So the exact time when the two hands overlap can be written as:the hour (n) + the time taken during that hour (T)(or)n + n/11where n = 0, 1, 2, 3, ..., 24.AM12:001:052:113:164:225:276:337:388:449:4910:55PM12:001:052:113:164:225:276:337:388:449:4910:5522 is correct.I would propose that the hands always overlap, as they're both attached at the center of the dial.If you didn't want to be facetious (or, at least, less facetious), you would still have to ask how many hands were on the clock. It may have a second hand, for example, or be digital (no hands at all).22 times a day if you only count the minute and hour hands overlapping. The approximate times are listed below. (For the precise times, see the related question.)2 times a day if you only count when all three hands overlap. This occurs at midnight and noon.am12:00 1:052:113:164:225:276:337:388:449:4910:55pm12:00 1:052:113:164:225:276:337:388:449:4910:55A really simple way to see this is to imagine that the two hands are racing each other around a track. Every time the minute hand 'laps' the hour hand, we have the overlaps we want.So, we can say that the number of laps completed by the minute hand every T hours, Lm = T laps. Since there are 12hours in a full rotation of the hour hand, that hand only rotates Lh = T/12 laps.In order for the first 'lapping' to occur, the minute hand must do one more lap than the hour hand: Lm = Lh +1, so we get T = T/12 + 1 and that tells us that the first overlap happens after T = (12/11) hours. Similarly, the 2nd lapping will occur when Lm = Lh + 2.In general, the 'Nth' lapping will occur when Lm = Lh +N, which means every N*(12/11) hours (for N = 0,1,2,3...). In other words, it will happen approximately every 1hr5mins27secs, starting at 00:00. In 24hours, this occurs a total of 24/(12/11) = 22 times.======================================================================================So we are looking at two rotating hands. Ultimately, its just the angles we care about. Let θH represent the angle of the hours hand and θM represent the angle of the minutes hand. You could also introduce the seconds hand but that makes the problem more complicated. For now, lets assume the question only cares about the minute and hour hands. Initially we might think we are looking for:θH=θMBut this doesn't take into account that if one hand has "gone around" a few times, its angle will be different from a hand in the same position that hasn't "gone around" the same number of times. So we have to modify our goal. Instead we let the angles differ by an integer multiple of 2π (360°). Let us call this arbitrary integer z. Now our condition is:θM-θH=2πzYou could subtract the two angles in either order but the reason I chose to subtract hours from minutes is because it will result in positive integers which is just simpler. The minute hand goes around more times, thus its angle is bigger, thus this order of subtraction is positive. Now we have to find out how these angles depend upon the time. Let us call our time t and measure it in hours. I omit units for simplicity. The hour hand goes around a full rotation (2π) once every 12 hours. So:θH=(2π/12) tFor those more versed in mathematics, 2π/12 is the "angular frequency" for the hour hand (usually denoted by ω).Similarly the minute hand goes around a full rotation (2π) once every hour. So:θM=2π tPlugging back in:θM-θH=2πz2π t - (2π/12) t = 2πzt - t/12 = z(11/12) t = zNow we are ready to solve. The two hands overlap at every solution of this equation, so we want to know the number of solutions of this equation. But remember, we want to know how many times this happens in a single day, so t cant be bigger than 24 (remember we are measuring t in hours), and technically no smaller than 0 (assuming we start our clock at 0 hours). Since t and z are proportional, each solution for z corresponds to exactly one solution for t, and accordingly exactly one solution of the equation.Also, remember than z must be an integer. So if we wanted all the times we would just let z go from 0 (when t=0) up and solve for t and stop as soon as we passed t=24. Then of course we'd have to convert that into hour and minute format. However, we only care about the number of times this happens. So we can notice that as t increases, z is just keeping track of how many times the two hands have overlapped. When z=0 we get the first time, when z=1 we get the second time, and so on. Since t and z are directly proportional, t increases with z, thus z represents the number of times the hands have overlapped up until time t minus 1 (and starting from t=0). Since we don't want t to go past 24, we plug in 24 and solve for z which will tell us how many times this event has occurred from t=0 to t=24 (one day).(11/12)*24 = z22 = zSo this happens 22 times in a day. Technically this has 23 solutions (0 through 22) but the last one is for t=24 which has begun the next day. If we don't count that solution we are left with 22.■If we want the second hand to overlap as well, we have to go a bit further. First we note that the second hand makes a full rotation once every minute, thus 60 times an hour. From this we have:θS=(2π*60) tWe want the second and hour hands to overlap AND the minute and second hands to overlap. Those conditions can be summarized as follows, where x and y are positive integers:θS-θM=2πxθS-θH=2πyPlugging in our functions of t for the θ's and solving for t we are left with:t=x/59t=12y/719We want our integers x and y to produce the same time (making all hands overlap at that time). So we want to set the two equations equal. Simplifying, we get;x=708y/719708 and 719 are coprime (719 is prime and 708 is decomposed into 2^2*3*59). In fact 708y and 719 are coprime except for when y is an integer multiple of 719. Thus 708y/719 can only be reduced when y=719k for some integer k. In this case we have:x=708kThe first solution is when k=0. Then x=0 and t=0 corresponding to midnight. The next solution is k=1. Then x=708 and t=12 corresponding to noon. The next solution is k=2 but this corresponds to t=24 which is (midnight for) the next day and due to the direct proportionality of t and k, every k from here on up will produce t's higher than 24.In summary, all three hands only overlap twice a day: at noon and midnight. ■All of this assumes that the hands sweep continuously. So the math is more(?) complicated for those with fake Rolex's (or any ticking handed clocks).Starting at 00:00 (midnight) with overlapping hands. At noon the hands have overlapped an additional 12 times and at midnight another additional 12 so including the starting and finishing midnight overlaps, 25 times.
How much is a local call dialing 0844 first?
In the UK, dialing an 0844 number is almost always more expensive than a local call, so there is no reason to use an 0844 number for a local call.
What is an 0844 number?
An 0844 number is a number that costs the caller additional per-minute rates and puts more money in the business's pocket.
What is the phone number of Blemain Finance Limited?
If you need to contact Blemain Finance Limited their number is 0844 873 4120. You can also contact their team via 0844 873 4293 or their underwriting relationship managers through 0844 873 4127.
In what country are 0844 phone numbers used?
The country that uses the 0844 phone number as its code is the United Kingdom. This code provides phone services to businesses within the United Kingdom.
What is British Airway's phone number?
Customer relations : 0844 493 0787
What is the contact for Cineworld Bradford?
You can call the number, 0871-200-2000. This is a telephone booking line. Or you can call the number, 0844-815-7747 for costumer service. Also, you can call this number, 0844-372-3445, for an unlimited Card helpline.
What is the contact information for Phones4You?
Phones4You have several methods of contact. The quickest method is to call them on 0844 871 2231. If you need a specific department they will be able to route you from that number. You can also email them or write to them and details for those contact methods are on their website.
What is the common deflection of a M777?
the common deflection for the m777 and m198 is 3200 mil, i am a 0844 (fdc)
What is Wiz Khalifa cell number?
254 466 0844
What is the ISBN of The White Rose novel?
The ISBN of The White Rose - novel - is 0-8125-0844-0.
Would someone inform you How do you contact the Bereavement Office of Barclays Bank?
Dial 0844 633 0888
What is the telephone number of Equifax UK?
0844 414 6073 0870 060 1414 Try either of these
