Answer:
3960.4 bacteria
Step-by-step explanation:
The formula to solve the above question is given as:
P(t) = Po (2) ^t/k
P(t) = Population after time t = ?
Po = Initial population = 650 bacteria
t = Time in days = 7.3 days
k = doubling time = 2.8 days
P(t) = 650 × (2)^7.3/2.8
P(t) = 650 × 2^2.6071428571
P(t) = 650 × 6.0929582599
P(t) = 3960.4228689 bacteria.
Approximately = 3960.4 bacteria
Therefore, the number of bacteria the researcher will have after 7.3 days if they started with 650 bacteria is 3960.4 bacteria.
Answer:
864
Step-by-step explanation:
144x6
Answer:
- <u><em>The solution to f(x) = s(x) is x = 2012. </em></u>
Explanation:
<u>Rewrite the table and the choices for better understanding:</u>
<em>Enrollment at a Technical School </em>
Year (x) First Year f(x) Second Year s(x)
2009 785 756
2010 740 785
2011 690 710
2012 732 732
2013 781 755
Which of the following statements is true based on the data in the table?
- The solution to f(x) = s(x) is x = 2012.
- The solution to f(x) = s(x) is x = 732.
- The solution to f(x) = s(x) is x = 2011.
- The solution to f(x) = s(x) is x = 710.
<h2>Solution</h2>
The question requires to find which of the options represents the solution to f(x) = s(x).
That means that you must find the year (value of x) for which the two functions, the enrollment the first year, f(x), and the enrollment the second year s(x), are equal.
The table shows that the values of f(x) and s(x) are equal to 732 (students enrolled) in the year 2012,<em> x = 2012. </em>
Thus, the correct choice is the third one:
- The solution to f(x) = s(x) is x = 2012.
This question is Incomplete
Complete Question
Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.)
Answer:
0.7762
Step-by-step explanation:
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
Population mean = 6.20 bl/s
Standard deviation = 1.58 bl/s.
x = 5 bl/s
z = 5 - 6.20/1.58
z = -0.75949
The probability that an ant's speed in light traffic is faster than 5 bl/s is P( x > 5)
Probability value from Z-Table:
P(x<5) = 0.22378
P(x>5) = 1 - P(x<5)
= 1 - 22378
= 0.77622
Approximately to 4 decimal places = 0.7762
The probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7762
Not sure about the question, but answer is:
3/5* 2/5 = 6/25=0.24 =24%