So they get $2 on day 0, $2 on day 2, $3 on day 4, $5 on day 5, $2 on day 6, $1 on day 7, $2 on day 8, $4 on day 10, $1 on day 11, $2 on day 12 and $1 on day 15.
$2 appears 5 times
$1 appears 3 times
$4 appears once and $5 appears once
The median would be ($2*5 + $1*3 + $4 + $5)/(5+3+1+1) = $22/10 = $2.2
The range of the allowances is highest - lowest = $5-$1 = $4
Edit: It might also have a range of $5 and a median of $1.37 because the lowest it goes to is $0 on some days though. Try both.
Answer: 10.5 is my guess
Step-by-step explanation:
1. 30% = 8.4
100 x
cross multiply
30(x) = 840
divide both sides by 30 = 28
2. 37.5% = y
100 28
cross multiply
1050 = 100y
divide both sides by 100 = 10.5
Answer:
- 31 pencils
- 38 erasers
- 44 sharpeners
Step-by-step explanation:
The number of packets is the greatest common divisor of the given numbers of pencils, erasers, and sharpeners.
It can be helpful to look at the differences between these numbers:
748 -646 = 102
646 -527 = 119
The difference of these differences is 17, suggesting that will be the number of packets possible.
527 = 17 × 31
646 = 17 × 38
748 = 17 × 44
The numbers 31, 38, and 44 are relatively prime (31 is actually prime), so there can be no greater number of packets than 17.
There will be 31 pencils, 38 erasers, and 44 sharpeners in each of the 17 packets.
_____
We may have worked the wrong problem. The way it is worded, the <em>maximum</em> number of items in each packet will be 527 pencils, 646 erasers, and 748 sharpeners in one (1) packet. The <em>minimum</em> number of items in each packet will be the number that corresponds to the maximum number of packets. Since 17 is the maximum number of packets, each packet's contents are as described above.
17 is the only common factor of the given numbers, so will be the number of groups (plural) into which the items can be arranged.
