3. in california (ca), standard license plates are currently of the form 1 number–3 letters–3 numbers.
assume the numbers are 0—9 and the letters are a–z.
a. in theory, how many different possible standard ca license plates are there, assuming we can repeat
letters and numbers?
b. how many different possible standard ca license plates are there if we are not allowed to repeat any
letters or numbers?
c. for part (b), did you use permutations or combinations to carry out the calculation? explain how you know.

a) 175,760,000

b) 78,624,000

c) Permutation

Step-by-step explanation:

Data provided in the question:

standard license plates are currently of the form 1 number–3 letters–3 numbers.

The number plate will have following blanks

(_) (_ _ _) (_ _ _)

Now,

For the first bracket we have 10 choices i.e from 0 - 9

for the second bracket (i.e 3 blanks for letters ) we have 26 choices {as we have 26 alphabets }

for the third bracket  (i.e 3 blanks for number ) we have 10 choices for every blank in the third bracket

Therefore,

For a) when we can repeat  letters and numbers

Number of possible ways = ( 10 ) × (26 × 26 × 26) × ( 10 × 10 × 10 )

= 175,760,000

For b)  if we are not allowed to repeat any  letters or numbers

Number of possible ways = ( 10 ) × (26 × 25 × 24 ) × ( 9 × 8 × 7 )

=  78,624,000

c) For the part b) we used the permutation for the calculation because in the given case the order of the number and letters matter

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

853.3

step-by-step explanation: