the insurance premium for you car is £400. your offered a 4.5% discount if you pay it other the phone or using the internet. what is the new premium? (hint: it is a discount so you should subract dont add)
2007-02-16 21:26:20 · 15 answers · asked by kelly g 1 in Science & Mathematics ➔ Mathematics
=(4.5%)(£400)
~>
=0.045(400)
~>£18 (discount)
£400(original price) - £18(discount) = £382
2007-02-16 21:59:46 · answer #1 · answered by jAmPr0tzz 2 · 0⤊ 0⤋
Discount = 4.5 % of £400
= 4.5/100 x £400
= £ 4.5 x 4 = £18
New premium = £400 - £18 = £382
2007-02-17 05:33:11 · answer #2 · answered by Como 7 · 1⤊ 0⤋
If you pay using internet and get 4.5% discount then
4.5 % of £400=18
Total Discount=400-18=£382 Answer
2007-02-17 05:50:23 · answer #3 · answered by APNim 2 · 0⤊ 0⤋
1% of £400 = £4..so 0.5% =£2 and 4% = £16
4.5% of £400 = £18...discount.(£16+£2)
£400-£18 = £382.
Hints not required ...this is a simple problem..we did these in primary school.
2007-02-17 05:31:48 · answer #4 · answered by Anonymous · 0⤊ 0⤋
400 = 100%
4 = 1%
100 - 4.5 = 95.5
95.5 * 4 = 382
new premiun is £382
2007-02-17 12:35:24 · answer #5 · answered by crazychick_woman 4 · 0⤊ 0⤋
£382
2007-02-17 13:35:47 · answer #6 · answered by Luke W 2 · 0⤊ 0⤋
£382.... 1% = £4 so 4% = £16 + 0.5% ... £16 + £2 = £18.
2007-02-17 09:02:58 · answer #7 · answered by Anonymous · 0⤊ 0⤋
10% of 400 is 40
so u take 40 times 4 equals to 160 and add half of 40 since it is 0.5 % and u get 180 so u take 400 and minus 180 and u get 220 pounds
2007-02-17 05:29:36 · answer #8 · answered by Anonymous · 0⤊ 2⤋
A=P(1-r)^1 r=%/100
A=400(1-0.045)^1
A=400(0.955)
A=382
or
A=P-P(%/100)
A=400-400(4.5/100)
A=400-400(0.045)
A=400-18
A=382
2007-02-17 06:51:33 · answer #9 · answered by rA_0215 2 · 0⤊ 0⤋
Dont post Your maths homework along with the hint your teacher gave..
Yet i think some people have got time to get you answers..
I suggest you learn the concept and give yourself some practice , because yahoo answers are not allowed during Exams.
2007-02-17 05:37:33 · answer #10 · answered by Nancy Maxilus 2 · 0⤊ 1⤋