let x be how much longer she can run the race and still beat her previous time
55(min) + x(min) < 1(hr) + 10(min)
55(min) + x(min) < 60(min) + 10(min)
55 + x < 60 + 10
x < 60 + 10 - 55
x < 15(min)
Answer:
2
Step-by-step explanation:
Given the question :
Serena wants to create snack bags for a trip she is going on. She has 6 granola bars and 10 pieces of dried fruit. If the snack bags should be identified without any food leftover, what is the greatest number of snack bags Serena can make?
Number of granolas = 6
Number of dried fruits = 10
Since the snackbag is to be designed in such a way that there should be no food leftover, the greatest number of snack bags Serena can make could be obtained by getting the highest common factor of (6 and 10)
____6____10
2___3____5
Here, the highest common factor of 6 and 10 is 2
Hence, the greatest number of snack bags she can make is 2.
The mean is 10.
To find the mean you add up all the numbers and divide by how many sets of numbers there were.
8+10+11+8+13=50
50/5=10.
Answer:
82 ducky grouses, 41 trumpeter swans, 26 sand hill cranes, 164 chikadees
Step-by-step explanation:
6(2y-4)+p
If you distribute you can get
2y-24+p ;)
