Answer:
The amount of ⛅ sunlight .
Explanation:
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Answer:
In order to find average speed during each interval, we need to divide the distance during those intervals with the period of time. So, for the first interval (day 0 to day 2) hawksbill started from 0 and reached 10 kilometers by the end of the second day. That means that it crossed 10 kilometers in 2 days, so the average speed is 10/2 which is 5 km/day. Similarly, we can calculate speed for other intervals:
• day 2 - day 3: it went from 10 to 12 km in one day, which means it crossed 2 km in one day, so the average speed is 2/1 = 2 km/day
• day 3 - day 4: at the end of the third day it reached 12 km and at the end of the day 4 it remained at 12 km. That means the hawksbill wasn't moving in that interval so the speed was 0
• day 4 - day 5: it went from 12 km to 18 km, which means it crossed 18-12=6 km in one day, so the average speed is 6/1=6 km/day
• day 5 - day 6: it went from 18 to 24 km, which means it crossed 24-18=6 km in one day, so the speed was 6/1=6 km/day
So, to summarize, during the first interval turtle was moving with average speed of 5 km/day, then 2 km/day, in the third interval it wasn't moving and in the last two intervals, it moved in average speed of 6 km/day.
A. Consumers cannot produce their own food source so they have to eat other animals.
C. Rhizobia bacteria "prepare" the atmospheric nitrogen and make it useful for the plant.
Hope it helped,
Happy homework/ study/ exam!
Answer:
25%
Explanation:
Let's assume that the recessive allele "p" imparts diseased conditions in the homozygous genotypes. The genotype of each of the carrier parents would be "Pp". A cross between Pp and Pp would produce progeny in the following phenotype ratio=
Pp x Pp= 3/4 Normal : 1/4 Affected.
Therefore, there are 1/4 or 25% chances for this couple to have a child with PKU.
