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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:
x ≈ 83.533
y = 9105.13
Step-by-step explanation:
Step 1: Substitution
109x = 79x + 2506
30x = 2506
x = 83.533
Step 2: Plug <em>x</em> in
y = 109(83.533)
y = 9105.13
Graphically:
Horizontal means X axiz. so 6 units on the x axis. The points are always placed (x,y) which is A. (1,6) and (7,6)
Step-by-step explanation:
the introduction of a fraction tells us that we are dealing with multiplications, and therefore a geometric sequence (where every new term is created by multiplying the previous term by a constant factor, the ratio r).
I think your teacher made a mistake, or you made one when typing the question in here.
there is no factor r that creates
15×r = 9
and
9×r = 5/27
it would mean that
15 × r² = 5/27
r² = 5/27 / 15 = 5/27 × 1/15 = 5/405 = 1/81
r = 1/9
but 15 × 1/9 = 5 × 1/3 = 5/3 is NOT 9
and 9 × 1/9 = 9/9 = 1 is NOT 5/27
so, this can't be right.
on the other hand
15 × r = 9
r = 9/15 = 3/5
and then
9 × 3/5 = 27/5
so, either the sequence should have been
15, 5/3, 5/27
or (and I suspect this to be true)
15, 9, 27/5
under that assumption we have
s1 = 15
r = 3/5
sn = sn-1 × r = s1 × r^(n-1) = 15 × (3/5)^(n-1)
s10 = 15 × (3/5)⁹ = 15 × 19683/1953125 =
= 3 × 19683/390625 = 59049/390625 =
= 0.15116544 ≈ 0.151
