Answer: The correct option is 5.7/20 = x/200
Step-by-step explanation: Please refer to the diagram attached for more details.
From the 7-column by 2-row table drawn, you can determine that the mean number of catfish as given in the question is derived as follows;
Mean = Summation of observed data / Number of observations
Mean = (3 + 5 + 7 + 5 + 6 + 8)/6
Mean = 34/6
Mean = 5.6667
Mean ≈ 5.7
Also the mean number of all fish is derived as follows;
Mean = Summation of observed data / Number of observations
Mean = (20 + 20 + 20 + 20 + 20 + 20)/6
Mean = 120/6
Mean = 20
Using relative frequency, we can confidently predict the number of catfish population in the pond. This simply means, if the mean number of catfish is 5.7 and the mean number of all fish is 20, then the relative frequency would be 5.7 over 20 of the total fish population.
Hence, if there is a total of 200 fish in the pond then to predict the proportion which would be catfish can be expressed as follows;
StartFraction 5.7 over 20 EndFraction = StartFraction x over 200 EndFraction.
OR better expressed as;
5.7/20 = x/200
Answer:
The distance between the two airplanes (to the nearest mile) is 1058 miles.
Step-by-step explanation:
An airplane A is at a location 800 miles due west of city X. So AX = 800 miles.
Another airplane is at a distance of 1,200 miles southwest of city X. So BX = 1200 miles.
The angle at city X created by the paths of the two planes moving away from city X measures 60°. So angle ∠AXB = 60°.
In triangle ΔAXB, AX = 800 miles, BX = 1200 miles, ∠AXB = 60°.
Using law of cosines:-
AB² = AX² + BX² - 2 * AX * BX * cos(∠AXB).
AB² = 800² + 1200² - 2 * 800 * 1200 * cos(60°).
AB² = 640000 + 1440000 - 2 * 960000 * 1/2
AB² = 2080000 - 960000
AB² = 1120000
AB = √(1120000) = 1058.300524
Hence, the distance between the two airplanes (to the nearest mile) is 1058 miles.
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C Would Be Correct I Belive
Answer:
4700 ft
Step-by-step explanation:
Find the area (l x w)
94 x 50 = 4700 feet
DON'T FORGET UNITS