_.map(randRange(1e3,16e3,randRange(3,7)),function(e){var r=e/1e3,n=randRange(1,2);return[roundTo(n,r),n]})
sum($.map(VALUES,function(e){return round(e[0])}))
randFromArray([["jogging","ran",new Plural(function(e){return $.ngettext({lang:"es-ES",messages:["milla","millas"]},e)})],["riding a bicycle","traveled",new Plural(function(e){return $.ngettext({lang:"es-ES",messages:["milla","millas"]},e)})],["collecting mushrooms","collected",new Plural(function(e){return $.ngettext("pound","pounds",e)})],["painting a roof","painted",new Plural(function(e){return $.ngettext("square foot","square feet",e)})],["harvesting grain","harvested",new Plural(function(e){return $.ngettext("tonne","tonnes",e)})]])
0
function(){var e=0;return $.each(VALUES,function(r,n){e+=roundTo(n[1],n[0])}),roundTo(2,e)}()
person(1) has been TYPES[0] every day for the last VALUES.length days.
Given that person(1) TYPES[1] the following amounts (in plural_form(TYPES[2])) each day, estimate how much he TYPES[1] in total by first rounding each amount to the nearest whole number and then summing all the rounded amounts.
Given that person(1) TYPES[1] the following amounts (in plural_form(TYPES[2])) each day, estimate how much she TYPES[1] in total by first rounding each amount to the nearest whole number and then summing all the rounded amounts.
i+1
localeToFixed(pair[0], pair[1])
Note that if you first sum the exact amounts (localeToFixed(SUM, 2)) and then round, you'll get round(SUM). For fun, pay attention to how this number compares with your estimate.
ANSWER plural_form(TYPES[2], ANSWER)
Round all the numbers to the nearest plural_form(TYPES[2], 1) and add them up.
i+1
localeToFixed(pair[0], pair[1])
round(pair[0])
ACC += round(pair[0])
Therefore the estimate is ANSWER plural_form(TYPES[2], ANSWER).