I believe it is C hope this helps
Answer: 32x^1/2y^2
Step-by-step explanation:
Multiply 64 times 1
64xy^4/2
Then divide 64 by 2
32x1/2y^2
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
Answer:
g ≈ 34.7 in
Step-by-step explanation:
The law of sines is useful for this:
f/sin(F) = g/sin(G)
Multiplying by sin(G), we have ...
g = f·sin(G)/sin(F) = (61 in)·sin(34°)/sin(79°)
g ≈ 34.7 in
Answer:
14
Step-by-step explanation:
f(x)=3x²+1
f(2) = 3 x 2² + 1 = 13
g(x) =1-x
g(2) = 1 - 2 = -1
(f-g)(2) f(2) - g(2) = 13 - (-1) = 14
