Partial credit, 8.3.22-T Question Help A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.05 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?
77
93
50
76
49
33
67
70
72
60
72
82
100
100
74

Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?
A.
H0: μ=60 seconds
H1: μ>60 seconds
B.
H0: μ≠60 seconds
H1: μ=60 seconds
C.
H0: μ=60 seconds
H1: μ≠60 seconds
Your answer is correct. D.
H0: μ=60 seconds
H1: μ<60 seconds
Determine the test statistic.
nothing (Round to two decimal places as needed.)

  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).

sorry i just need to anwers this to get points

step-by-step explanation:

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.

the answer to this is a. a^2 + b^2 = c^2