9/11 =?/22 what is the answer
2 answers:
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The linear relationship which exists between the change in weight and number of days of hibernation in a hedgedog can be modeled using the equation y = - 0.03 + 0
- Change in weight per day of hibernation = - 0.03
- Change in weight for 115 days = - 3.45 ounces
A linear model can be created using the data in the table given using a regression calculator or excel ;
The linear model obtained in the form y = m + bx is :
- y = - 0.03x + 0
- y = change in weight
- x = number of days of hibernation
- Intercept = 0
- Slope = -0.03
- The slope value of the function gives the change in weight value per number of days of hibernation ; which is -0.03.
- Using the regression equation, substitute, the vlaue of x = 115
- -0.03(115) + 0 = - 3.45
Therefore, the change in weight after 115 days of hibernation is - 3.45 ounces.
Learn more :brainly.com/question/18405415
Answer:
- h = -16t^2 + 73t + 5
- h = -16t^2 + 5
- h = -4.9t^2 + 73t + 1.5
- h = -4.9t^2 + 1.5
Step-by-step explanation:
The general equation we use for ballistic motion is ...
where g is the acceleration due to gravity, v₀ is the initial upward velocity, and h₀ is the initial height.
The values of g commonly used are -32 ft/s², or -4.9 m/s². Units are consistent when the former is used with velocity in ft/s and height in feet. The latter is used when velocity is in m/s, and height is in meters.
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Dwayne throws a ball with an initial velocity of 73 feet/second. Dwayne holds the ball 5 feet off the ground before throwing it. (h = -16t^2 + 73t + 5)
A watermelon falls from a height of 5 feet to splatter on the ground below. (h = -16t^2 + 5)
Marcella shoots a foam dart at a target. She holds the dart gun 1.5 meters off the ground before firing. The dart leaves the gun traveling 73 meters/second. (h = -4.9t^2 + 73t + 1.5)
Greg drops a life raft off the side of a boat 1.5 meters above the water. (h = -4.9t^2 + 1.5)
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<em>Additional comment on these scenarios</em>
The dart and ball are described as being launched at 73 units per second. Generally, we expect launches of these kinds of objects to have a significant horizontal component. However, these equations are only for <em>vertical</em> motion, so we must assume the launches are <em>straight up</em> (or that the up-directed component of motion is 73 units/second).