Answer:1)Number of different license plates can be made =17576×100×260=456976000
2)Probability of getting a plate with abc123 or 123 abc=0.0000004
Given that :-A license plate has six characters.
As three characters are letters ∴ ways of these 3 letters into the plate with repetition = 26×26×26=17576 ways
and two characters are numbers (0-9 - total 10 characters)∴ways of these 2 numbers into the plate with repetition=10×10 =100 ways
and one character is a letter or a number=26×10=260
So number of different license plates can be made =17576×100×260=456976000
Now Probability of getting a plate with abc123 or 123 abc
=P(abc123 or 123abc)=P(abc123)+P(123abc)-P(abc123)×P(123abc)=1/456976000+1/456976000-1/456976000×1/456976000
=1/456976000(1+1-1/456976000)
=1/456976000(2-0.0000002)=2/456976000=0.0000004
Probability of getting a plate with abc123 or 123 abc=0.0000004
Makaylah: 7371 / 250
Keisha: 4731
/ 250
Kelviona: 8871
/250
Janell: 2217
/125
(i dont know the divison part)
By using proportions, We expect 150 defects out of the 10,000 cars if the 3 defects out of the 200 cars sampled is representative of all 10,000 cars.
Option A is correct.
First, let's set up the proportions.
A proportion is an equation in which two ratios are set equal to each other.
We have 3/200 cars with defects and we want to know X for X/10000.
So 3/200 = X/10000
Now solve for X.
3/200 = X/10000
X = 
X = 150
So the answer is: We expect 150 defects out of the 10,000 cars if the 3 defects out of the 200 cars sampled is representative of all 10,000 cars.
Option A is correct.
Find out more information about proportions here
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Answer:
The probability that the result is positive is P=0.04475=4.475%.
Step-by-step explanation:
We have the events:
D: disease present
ND: disease not present
P: test positive
F: test false
Then, the information we have is:
P(D)=0.005
P(P | D)=0.99
P(P | ND)=0.04
The total amount of positive test are the sum of the positive when the disease is present and the false positives (positive tests when the disease is not present).

The probability that the result is positive is P=0.04475.