Step-by-step explanation:
2/3 is right and ok then thank
1. True because a straight line is a function
2. IDK because y is f(x) but it is not written there so I am guessing True.
Answer:
Period = 8π
Phase shift = 2π/3
Amplitude = 1/2
Vertical shift = 2
Step-by-step explanation:
y = A sin((2π/T)θ − B) + C
where A is the amplitude,
T is the period,
B is the phase shift,
and C is the vertical shift.
y = ½ sin(¼θ − 2π/3) + 2
So A = 1/2, T = 8π, B = 2π/3, and C = 2.
Answer:
D - Jane is trying to determine the distance to school
Step-by-step explanation:
Option D is the most acceptable time to use approximate numbers and estimate how far it is to school. All of the other options, exact numbers are needed.
The correct answer is the first:
The interquartile range of Mitchell's data is greater than the interquartile range of Luke's data.
Since the interquartile range (RIC) is equal to:
RIC = Q3-Q1
Of the gracifa can be observed quer for the case of Mitchell:
RIC = 34-25 = 9
While for Luke's case:
RIC = 36-28 = 8.
Therefore Mitchell's interquartile range (RIC) is greater than Luke's.