Answer:
$66,092.50
Step-by-step explanation:
In 1988, the car cost $17,000
Annual inflation over this period was 3.54% from 1988 to 2017
After adjusted for inflation, $17,000.00 in 1978 is equal to $66,092.50 in 2017
Therefore, the cost of that car in 2017 is closest to $66,092.50
Plug in the values of p and q where you see them in the equation:
-(2+4)2 / (-6) - Distribute the -1
(-2-4)2 / (-6) - Distribute the 2
(-4-8) / (-6) - Subtract what's inside the parenthesis
(-12) / (-6) - Divide
The answer is 2
bearing in mind, "t" is in years, 6months is just 1/2 year
solve for "r", you'd get a decimal amount, multiply times 100, to get the percentage form
Answer:
The probability that the town has 30 or fewer residents with the illness = 0.00052.
Step-by-step explanation:
So, we have the following set of data or information or parameters given from the question above and they are; the number of people living in that particular society/community/town = 74,000 residents and the proportion of people that the diseases affected = .000215.
The first step to do is to determine the expected number of people with disease. Thus, the expected number of people with disease = 74,000 × .000215 = 15.91.
Hence, the probability that the town has 30 or fewer residents with the illness = 1.23 × 10^-7 × 15.91^30/ 2.65253 × 10^-32 = 0.00052.
Note the formula used in the calculating the probability that the town has 30 or fewer residents with the illness = e^-λ × λ^x/ x!
