Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

c. the fish and wildlife agency measured the size of pacific salmon for…

Question

c. the fish and wildlife agency measured the size of pacific salmon for 1 year and recorded the average weight for each species.
salmon species: king, sockeye, silver, chum, humpback
average weight (lbs): 15, 8, 12, 15, 5
d. the national oceanic and atmospheric administration measured the pressure of the atmosphere at various altitudes.
altitude (m): 0, 2750, 5486, 8376
atmos. pressure (atm): 1.000, 0.750, 0.500, 0.333
altitude (m): 16132, 30901, 48467
atmos. pressure (atm): 0.100, 0.010, 0.001

Explanation:

Response
Part c: Average Weight of Salmon Species

Step1: Identify the weights

The weights for each species are: King = 15 lbs, Sockeye = 8 lbs, Silver = 12 lbs, Chum = 15 lbs, Humpback = 5 lbs.

Step2: Sum the weights

Sum = \( 15 + 8 + 12 + 15 + 5 \)
\( = 15+8 = 23 \); \( 23+12 = 35 \); \( 35+15 = 50 \); \( 50+5 = 55 \)

Step3: Calculate the average

Number of species = 5. Average = \( \frac{55}{5} = 11 \)

Answer:

The average weight is 11 lbs.

Part d: (Assuming we need to analyze the pressure - altitude relationship, but since no specific question is given, if we assume finding a pattern or function, but based on the table, let's check the ratio)

Looking at the first table:

  • At altitude 0 m, pressure = 1.000 atm
  • At 2750 m, pressure = 0.750 atm (ratio \( \frac{0.750}{1.000}=0.75 \), altitude ratio \( \frac{2750}{0}= \) undefined, but maybe exponential? Wait, second table:

Altitude 16132 m, pressure 0.100 atm; 30901 m, 0.010 atm; 48467 m, 0.001 atm. Notice that as altitude increases by a factor (approx 30901/16132 ≈ 1.915, 48467/30901≈1.568), pressure decreases by a factor of 10 each time (0.100→0.010→0.001). The first table: 0→2750→5486→8376. Let's check the pressure ratios: 0.750/1.000 = 0.75; 0.500/0.750≈0.666; 0.333/0.500≈0.666. Maybe a linear or exponential decay. But since the question for part d is not clear, if we assume finding the missing pressure in the first table for altitude 8376 m (already given as 0.333), or maybe the relationship. But since the user's image has part c and d, and part c is solvable as above.

(Note: If the intended question for part d was different, please clarify. For now, part c is solved as above.)