Twenty ten thousandths, Four thousandths, Three hundredths,Five tens,Seven Ones
Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
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(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
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(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
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(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000
<span>2x + x = 12
=> x =12/3 =4
so, original number is 84. </span>
15.1 is close to 14 so then you have 14/7 which approximately = 2
Answer:
years worked: 14
at least 10 years: 73%
Step-by-step explanation:
The mean is found by adding the years of service and dividing by the number of employees. The total years of service is 417, so the average is ...
average years worked = 417/30 = 13.9 ≈ 14 . . . years
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The percentage of employees that have worked there at least 10 years is found by counting the number with 10 or more years of service and dividing that count by the total number of employees. The result is then expressed as a percentage.
(10 years or over)/(total number) = 22/30 = 0.73_3 (a repeating decimal) ≈ 73%
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Comment on the working
A spreadsheet can be helpful for this. It has a function that can calculate the mean for you. Sorting the years of service into order can make it trivially easy to count the number that are 10 or more, or you can write a function that will do the count for you. (Also, less than 10 means the years are a single digit. There are 8 single-digit numbers in your list.) The hard part is copying 30 numbers without error.
