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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
Answer:
95
Step-by-step explanation:
Average =
Sum
Count
=
332
4
= 83
<span>0 is a natural number, which is part of the rational numbers.</span>
Let V, be the rate in still water and let C = rate river current
If the boat is going :
upstream, its rate is V-C and if going
downstream, its rate is V+C,
But V = 5C, then
Upstream Rate: 5C - C = 4 C
Downstream rate: 5C+C = 6C
Time = distance/Rate, (or time = distance/speed) , then:
Upstream time 12/4C and
Downstream time: 12/.6C
Upstream time +downstream time:= 2h30 ' then:
12/4C + 12/.6C = 2.5 hours
3/C + 2/C = 5/2 (2.5 h = 5/2)
Reduce to same denominator :
5C = 10 and Rate of the current = 2 mi/h
