VWrite an integer that this statement is describing. Susan owes more than Peter but less than John. Peter owes $3 and John owes $5?
$4.00
VWrite an integer that this statement is describing. Susan owes more than Peter but less than John. Peter owes $3 and John owes $5?
$4.00
Is interest of less than $10 taxable?
"Even if you do not receive Form 1099-INT, you must still report *all* of your taxable interest income," is pretty cut and dried. Ya think? It does not say that interest of less than $10 isn't reportable, does it? It says *all*. (My emphasis, just in case you missed it.) FYI, just because you don't get a 1099-INT does not mean that the IRS wasn't notified of the amount. Truth is, they were, but the bank decided to save a few trees and a few hundred thousand dollars in postage by not sending out the ones that were not required by law. The reporting to the IRS is done electronically and the cost per item is less than a penny vs several dollars to print, stuff, and mail a single paper form. I used to work in the banking industry and the cost of printing and mailing 1099-INT forms is much higher than you'd imagine. Since they are a "low use" item, the cost to run them and mail them out is much higher than the monthly statements is. Additionally, many banks now use the "substitute form rule" and simply add a notice to the bottom of your last statement of the year showing the total interest paid. This suffices for the separate 1099-INT. Edit: Uhhh... Yeah, the IRS does explicitly address the issue: "Even if you do not receive Form 1099-INT, you must still report *ALL* of your taxable interest income." All means just that: ALL
He means there are little infinities and big infinities for example between the numbers 1 and 2 there are a infinity amount of numbers between them like 1.2, 1.22, 1.22222 And a lot more. But there will be a even bigger infinity amount of numbers between 2 and 3 or 1 and 3 and on and on because the numbers can easily get bigger creating loads and loads of little infinities between the numbers. That is why Hazel says she’s so happy that her little infinity involved Augustus Waters because there are so many people just like numbers and he managed to get into her little infinity to make her so happy that she got to share her little infinity with him.
In how many ways can a set of two positive integers less than 100 be chosen?
Sudha and Deepak have both given best approaches as a way to work with small corrections. Sudha - Make instances: a couple of of 3 times non-multiple of 3 and more than one of three times unique multiple of three Deepack - Compute the complement: find the likelihood of a NON-a couple of of 3 and subtract from 1 each have miscounted the number of positive integers lower than one hundred. Deepack is nearer, having counted 33 multiples of three and sixty six non-multiples of three. How many numbers altogether? However most effective Sudha has noticed that the obstacle requires that the 2 integers be specific Who will be the first to right? Deepack? Or Sudha? Or DetectiveKat? P.S. Tom can be shut, but has miscounted the quantity of non-multiples of three and in addition lost sight of the necessity of utilizing conditional chance. The chances of getting a more than one of three for A is 33/ninety nine. Having choosen that number, the likelihood of picking out another number for B which is not a multiple of 3 becomes?? Both Tom's numerator and denominator ought to be corrected. Our unknown man or woman with the seventh-grade mind has overpassed two things: 1) The made of two numbers below a hundred might be over one hundred. 2) The likelihood of each and every product will not be the equal. Due to the fact that 3=1&occasions;3 only but 6=1&instances;6=2×three, a made from "6" is twice as likely as a made of "3". P.P.S. Now we have a winner!! M3 using Deepak's approach wirtten with Sudha's flavor.
Let [math]S[/math] be the set of all integers between [math]100[/math] and [math]500[/math] that are divisible by [math]6[/math], and [math]E[/math] the set of those divisible by [math]8[/math]. The answer to your question is [math]|E\cup{S}|[/math]. By the inclusion-exclusion principle[math]\begin{align}\displaystyle |E\cup{S}|=|E|+|S|-|E\cap{S}|\end{align}\tag{1}[/math][math]|E|[/math]: the integers between [math]100[/math] and [math]500[/math] that are divisible by [math]8[/math] range from [math]8\cdot{13}[/math] to [math]8\cdot{62}[/math], so there are [math]62-13+1=50[/math] of them.[math]|S|[/math]: the integers divisible by [math]6[/math] range from [math]6\cdot{17}[/math] to [math]6\cdot{83}[/math], so there are [math]83-17+1=67[/math] of them.[math]|E\cap{S}|[/math]: the integer divisible by both [math]6[/math] and [math]8[/math] are those divisible by [math]\mathrm{lcm}(6,8)=24[/math]. They range from [math]24\cdot{5}[/math] to [math]24\cdot{20}[/math], so there are [math]20-5+1=16[/math] of them altogether.Plug these numbers in [math](1)[/math] and get[math]\begin{align}\displaystyle |E\cup{S}|=|E|+|S|-|E\cap{S}|=50+67-16=\boxed{101}\end{align}\tag*{}[/math]
My first job would be to find out how many times 7 goes into 200.200/7 = 28.57 But we can’t go over 200 so we will call it 28.But the 1st multiple of 7 is not included, because 7 is less than 10. However the 2nd multiple of 7 is included because 14 is greater than 10.So you want to add the 1st 28 multiples of 7 and then subtract 7.We know how to add consecutive integers, this is the triangular numbers [(n)(n+1)/2], but these are the multiples of 7, so we will take the 7 x [(n)(n+1)/2] where n = 28. But because the 1st multiple of 7 is missing, the final step will be to subtract 7.7 x [(28)(29)/2] - 7 = 2842 - 7 = 2835Another strategy for finding this arithmetic series sum is the formula:Sn = [(T1 + Tn)/2] x nWhere T1 is the 1st term 14 and Tn is the last term (28 x 7) 196 and n is 1 less than 28, or 27.Sn = [(14 + 196)/2] x 27 => 105 x 27 = 2835.
TO PROVE THAT: AC - AB < BCCONSTRUCTION: From the longer side AC, cut a segment AD = ABSo, now, we need to prove that AC - AD < BC=>TO PROVE first: that DC < BCPROOF:< ABD = < ADB = a ( by construction)< DBC = p, & < BDC = kSince, k = a + A ( exterior < of a triangle) ……. (1)p = a - C ( again by exterior < of a triangle) ….(2)By comparing (1) & (2)(a + A) > (a - C) , as all these angle variables >0=> k > p=> side opposite to k > side opposite to p=> BC > DC=> DC < BC=> AC - AD < BC=> AC - AB < BC[ hence proved]