25/30 bc 6x6=30 and 5x5=25
Answer:
3:40 p.m.
Step-by-step explanation:
Add 74 minutes to 1:25 p.m.; the improperly formed result witll be 1:99 p.m.; to express this properly, add 1 hour to this time and subtract 60 minutes from it:
2:39 p.m. (time at which Avery gets off the bus)
If Avery walked 61 minutes to get home and we want to know what time she arrived, we add 61 minutes to 2:39 p.m., obtaining 2:100 p.m., which must in turn be re-written as 3:40 p.m.
Avery arrived home at 3:40 p.m.
Note: another way in which to do this problem is to add 74 minutes and 61 minutes, obtaining 135 minutes, and then adding 135 minutes to 1:25 p.m. and making the necessary adjustments to the result:
1:25 p.m. + 135 minutes = 1:160 p.m., or (recognizing that 120 min = 2 hours)
3:40 p.m.
Answer:
Following are the solution to this question:
Step-by-step explanation:
Its laborer's journey through time was concentrated mostly on left as well as the amounts are mostly less than 4 min. There are quite a lot of significantly higher levels, even stretching to 150 min. A hypothetical worker is 15% likely to get a time difference longer than 45 min.
- They learn from such data that perhaps the distribution of data takes a very long time. Thus the distribution will look like an's progressive.
- Its probability is positive. Therefore some severe finding beyond the regular standards is included. The mean of findings is also higher than the mean.
- Its problem would be that 15% of people are more likely to have been traveling more than 45 min in a specific worker.
- Subsequently, the 85th level of the data is 45 minutes, meaning whether 85% of observations are decreased or equivalent to 45 minutes or 15% more than 45 minutes.
Answer:
I think the answer is 35%