I wrote answers in picture and I use Desmos app to solve this equations
a previous analysis of paper boxes showed that the standard deviation of their lengths is 15 millimeters. A packers wishes to fi
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Answer:
Question 9
(a) The distance, Jalaj walks in one day is 4.4 km
(b) The amount Jalaj raises after walking for 22 km at the end of the 5 days is $8
Question 10
(b) The difference between the largest and smallest areas is 2,400 m²
Step-by-step explanation:
Question 9
(a) The distance Jalaj walks in 5 days = 22 km
Whereby Jalaj walks equal distance every day, we have;
The distance, Jalaj walks in one day = 22 km/5 days = 4.4 km/day
The distance, Jalaj walks in one day = 4.4 km
(b) The amount he raises for every kilometer he walks = $1.60
The amount he raises after walking for 22 km at the end of the 5 days = 5 × $1.60 = $8
Question 10
(b) The given side length of the square = 120 meters to the nearest 10 meters
Therefore;
The maximum dimension for the side length of the square = 120 + 10/2 = 125
The largest possible area of the square, = 125 m × 125 m = 15,625 m²
The minimum dimension for the side length of the square = 120 m - 10 m/2 = 115 m
The smallest possible area of the square, = 115 m × 115 m = 13,225 m².
The difference between the largest and smallest areas, -
= 15,625 - 13,225 = 2,400 m².