Hello!
To find the interquartile range you subtract the lower quartile by the higher quartile
33 - 25 = 8
The answer is C)8
Hope this helps!
58÷40×100 which will give you 145 as the answer.
S(t) = 3.82 - 0.2t
Step-by-step explanation:
After 7 weeks, the ice is 2.42 meters thick. The ice loses 0.2 meters of thickness per week; this means on the 7th week, it has lost 7(0.2) = 1.4 meters of thickness.
This means the ice started at 2.42+1.4 = 3.82 meters thick.
Our function will start at the original thickness of the ice, 3.82 meters.
Since the ice is losing thickness, we will subtract; it loses at a rate of 0.2 meters per week (t), which gives us 0.2t. This is subtracted from the original, 3.82 meters, giving us
S(t) = 3.82 - 0.2t
Step-by-step explanation:
SO, OT is parallel to the vector a+2b
Answer:
Range: (-3,0,3,6,9)
Step-by-step explanation:
We need to evaluate each value in the domain of the function, so
f(-2)=3(-2)+3=-6+3=-3
f(-1)=3(-1)+3=-3+3=0
f(0)=3(0)+3=0+3=3
f(1)=3(1)+3=3+3=6
f(2)=3(2)+3=6+3=9
