Justin did pushups every weekday this week. He did 8 pushups on Monday, 14 push-ups on Tuesday, 18 push ups on Wenesday, 6 pushups on Thursday and 8 on friday. Find the median number of push-ups

The median number of push-ups is 8.

Step-by-step explanation:

We use the median to find the "middle" value of a sample, that is, the value separating the lower half from the upper half of the sample.

If the size of the sample is even

1,2,3,4

Both the upper half and the lower half should have the same size.

In this case, it is not possible to select only one value which respects this condition. So the mean between 2 and 3 = 2.5 is calculated. This value is not part of the sample.

If the size of the sample is odd

1,2,3,4,5

For a n-sized sample, with n odd, the median is the value at the position (n+1)/2. So her, it is 3, the value at the third position. In this case, it is always going to be an element at a position in the sample, that is, an element of the sample.

In this question:

Initially, we list the number of pushups sorted in an increasing way.

6,8,8,14,18

The size of the sample is 5.

The median value is at the position (5+1)/2 = 3.

So it is 8, which is the element at the 3rd position.

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)

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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

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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).