d) How many three-digit numbers greater than 330 can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?(

The question is incomplete. Here is the complete question.

(a) How many three-digit numbers can be formed from the digits 0,1,2,3,4,5 and 6, if each digit can be used only once?

(b) How many of these are odd numbers?

(c) How many are greater than 330?

(a) 180

              (b) 75

               (c) 105

Step-by-step explanation:

(a) In the group, there are 7 digits. A three-digit number can not start with zero, otherwise, it will be a 2-digit number. So:

For the hundreds position, there are 6 choices.

For the tens position, since the digit can be used only once, there are 6 choices.

For the unit position, there are 5 choices.

The total three-digit number formed is: 6*6*5 = 180

(b) To form an odd number, the unit position must be an odd digit, then:

unit position has 3 choices;

hundreds position has 5 choices;

tens position has 5 remaining choices.

The total three-digit odd number is: 3*5*5 = 75

(c) The number formed must be greater than 330, so:

If the number start with a 3, to be greater, there are 3 other choices (4, 5 and 6), so Tens position has 3 choices and Unit position has 5 choices.

Total number is: 3*5 = 15

Another possibility is the number starts with a digit bigger than 3 and so,  there are 3 choices.

Tens position has 6 choices;

Unit position has 5 choices;

Total possibilities are: 3*6*5 = 90

The total number of ways a three-digit number is greater than 330 is:

90 + 15 = 105

3030 ur welcome

step-by-step explanation:

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).