Given:
Pool swimming Ocean swimming Total
Age 30 and younger 228 54 282
Over 30 years old 142 185 327
total 370 239 609
Frequencies:
Pool swimming Ocean swimming Total
Age 30 and younger 228/609 = 0.37 54/609 = 0.09 282/609 = 0.46
Over 30 years old 142/609 = 0.23 185/609 = 0.30 327/609 = 0.54
<span>total 370/609 = 0.61 239/609 = 0.39 609/609 = 1.00</span>
<span>The relative frequency (rounded to the nearest hundredth) of a person over 30 years old who prefers to swim in the ocean is 0.30</span>
Answer: 5
Step-by-step explanation:20 times .25
Answer:
Inez worked for 2 hours 32 minutes more than Joe
Step-by-step explanation:
Here, we are to calculate the difference in the amount of time in which both of them spent working.
Mathematically, that would be the time spent by Inez minus the time spent by Joe
= 7 hrs 23 minutes - 4 hrs 51 minutes
This works like normal arithmetic subtraction;
Since we cannot subtract 51 minutes from 23 minutes, we need to borrow 1 hour ( 60 minutes) from 7 , and it becomes 6 while our minutes become (60 + 23 = 83 minutes).
So the difference here will be (6-4)hrs and (83-51) minutes = 2 hrs 32 minutes
Answer:
We drawn the graph.
Step-by-step explanation:
We have a line segment AB = 6cm and a point P outside that a line segment.
We want to construct a line PQ that is parallel to line segment AB.
If PQ is parallel to longer AB, this means that a line PQ is equidistant from line segment AB everywhere.
We use the geogebra.org site to draw it.
We drawn the graph.
Answer:
A p
Step-by-step explanation:
We have

Remeber that


Now, plug in
So now we plug in -1 for x and 5 for y

Simplify

A is the answer.