In a test of a printed circuit board using a random test pattern, an array of 14 bits is equally likely to be 0 or 1. Assume the bits are independent. (a) What is the probability that all bits are 1s? Round your answer to six decimal places (e. g. 98.765432).

The probability that all 14 bits are 1s = 0.000061

Step-by-step explanation:

Probability that a bit is 1 is p = 0.5

Probability that a bit isn't 1 is q = 1 - 0.5 = 0.5

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 14

x = Number of successes required = 14

p = probability of success = 0.5

q = probability of failure = 0.5

P(X = 14) = ¹⁴C₁₄ (0.5)¹⁴ (0.5)¹⁴⁻¹⁴

P(X = 14) = ¹⁴C₁₄ (0.5)¹⁴ (0.5)⁰

P(X = 14) = 1 (0.000061)(1) = 0.000061

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

the mean and mad

step-by-step explanation: