SinA=5root3/10
CosA=5/10
TanA=5root3/5=root3
SinC=5/10=1/2=0.5
CosC=5root3/10
Tanc=5/5root3=1/root3
Answer:
The answer is "Option A and Option B".
Step-by-step explanation:
In question 1:
In all cases, the entire population is measured so that the actual medium discrepancy could be measured as well as an interval of trust cannot be used.
This issue would be that she calculated the ages with all representatives of both classes, such that she measured a whole population. It's not necessary.
In question 2:
When the p-value is 0.042. At 90% trust and 92% trust level 11 (p-value below 0.10 and 0.08) are not included. however the biggest confidence level of 92%. Consequently, the largest trust level where the 11 is Not included in the trust interval is 92% trust.
16,500
60 minutes in an hour / 12 minutes = 5
5 * 3,300 = 16,500
The equivalent 90° CCW rotation maps (x, y) to (-y, x).
(4, -2) becomes (2, 4) after rotation.
Answer:
5:53
Step-by-step explanation:
The original time would be 37 minutes before 6:30 so you would subtract 37 minutes from 6;30. So 6:30 minus 30 would be 6 and you would have 7 left over so 6:00 minus 7 minutes would be 5:53