Acombination lock has a code consisting of 3 numbers, each of which can be 0 to 39, with numbers repeated. jillian says that there are only 120 possible codes. is jillian correct? if not, explain her error.

Jillian is not correct

Jillian has summed up the possible number of codes for each digit instead of multiplying it

Step-by-step explanation:

The lock has a code that consists of 3 number

It is given that the number in the lock code can be used between to

Thus, out of total set of numbers (i.e ), the numbers can be repeated.

This means for all three code numbers the opportunity of choosing a number is same  i.e. between  

Now, the first digit of the code can be any number between  

Like wise the second and third  digit of the code can be any number between  

Thus. the possible number of codes with repetition allowed are

Hence, Jillian is not correct

Jillian has summed up the possible number of codes for each digit instead of multiplying it

need more information

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?