A survey was done on the number of people in eachcar leaving a Shopping Centre. The results are summarised in the following table:
Number of People
in Each Car
Frequency
1
45
2
198
3
121
4
76
5
52
6
13
a. What is the Mode number of people in each car?
b. What is the Median number of people in each car?
c. What is the Mean number of people in each car? Give your answer to 1
decimal place.

a) Mode number = 2

b) Median  = 64.5  = 4 people in car as 0.5 in 64.5 is close to 76

              = 4 people

c) Mean number is  505/6 =   84.1666666667 then we see where 84.16 lies and see this lies with =  3 people

Step-by-step explanation:

45  is 1

198 is 2

121 is 3

76 is 4

52 is 5

13 is 6

Mode we look for the highest frequency and see 198 is the amount and 2 people in car is the subject, we give the subject and that is 2 people in car is the highest frequency and would be the mode.

Median is shown in the answer as we add up.

Mean we add up all frequency and divide by the amount of subjects = 6.

a) Mode number = 2

b) Median = 13, 45, 52, here 76, 121, 198    mid number between

             = (76-52 )/ 2  + (52)= 24 /2 + (52) = 12 + 52 + 0.5 if even number    

median  = 64.5  in a cumulative graph though which is not asked here it would be exactly half of 505 = 257.5 and interquartile we half again and add on 257.5 and show the range values interquartile as 1/2(252.5) and 1/2(252.5+ 505) = 126.25 as one value and 757.5 as the other by drawing a horizontal line from y axis to the points so vertical lines can represnt the cars median and interquartile.

c. Mean number is  505/6 =   84.1666666667 then we see where 84.16 lies and see this lies with 3 seats

For cumulative frequency we can relist numbers like this

45

45 +198 = 243

243 +  121 = 364

364+76 = 440

440 + 52 = 492

492+13 = 505 to plot graph.

But if there was group of measures or time etc then we would always plot the highest of each set for cumulative x axis and use the 2 values to see the ratio for histograms ie) 198/2 = 99 so that all values are read in equal proportions and 2 is a data below the graph not on the graph as 2 does not show as a box count just a name below on x axis. So we have to use the ratio as explained for histograms.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

1 and 0

step-by-step explanation: