in a certain examination 52 candidate offers biology,60 offers history,96 offers maths, if 21 offered both biology and history,22 offered maths and biology, and 16 offered maths and history, if 7 candidates offered all the subject. how many candidates were there for the examination​

There are 135 candidates.

Step-by-step explanation:

we know that:

52 offer biology

60 offers history

96 offers maths

21 offered both biology and history

22 offered maths and biology

16 offered maths and history

if 7 candidates offered all the subject.

We want to find the total number of candidates.

We start by adding the numbers for each particular exam:

(52 + 60 + 96)

Now, there are 21 that offered both biology and history, then these 21 are counted in the 52 for biology and in the 60 for history, then we need to subtract 21 (because we are counting it two times)

(52 + 60 + 96) - 21

The same happens for the 22 that offered maths and biology, and 16 that offered maths and history, then we get:

(52 + 60 + 96) - 21 - 22 - 16

And there are another 7 that offered for the 3 subjects, then we are counting these 7 ones 3 times, this means that we need to subtract 7 two times.

then we get:

(52 + 60 + 96) - 21 - 22 - 16 - 7 - 7

This is the total number of candidates, if we solve the equation we get:

candidates = (52 + 60 + 96) - 21 - 22 - 16 - 7 - 7 = 135

So we can conclude that there are 135 candidates.

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).