Answer:
Step-by-step explanation:
The power of a power rule is when power is raised to another power. (x^n)^m. when you do this rule, you must multiply the exponents so, x^nm or x^n*m.
The product rule for exponents is used when there are multiple terms that are raised by an exponent that have the same base and ae being multiplied. x^3 * x^2. To solve this, you keep the base (x) and add the two exponents together (3 and 2) so, x^3+2 = x^5.
Hope this helped (:
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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:
The correct option is (b).
Step-by-step explanation:
If X 
N (µ, σ²), then 
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z N (0, 1).
The distribution of these z-variate is known as the standard normal distribution.
The mean and standard deviation of the active minutes of students is:
<em>μ</em> = 60 minutes
<em>σ </em> = 12 minutes
Compute the <em>z</em>-score for the student being active 48 minutes as follows:
Thus, the <em>z</em>-score for the student being active 48 minutes is -1.0.
The correct option is (b).
