Since the ball achieves a peak height of 18 meters at 1 second, it will begin descending thereafter. Therefore, we anticipate that the height will drop below 18 m after 1.5 sec.
f(x) = –10x2 + 20x + 8
f(x) = –10(1.5^2) + 20(1.5) + 8
f(x) = 15.5 m
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Answer:
The resulting value is 2.381
Step-by-step explanation:
Using the information presented in the question, we will calculate the evidence supporting the professor's hypothesis
Given that:
x₁ = 74,
n₁ = 36
s₁ = 8
x₂ = 68
n₂ = 36
s₂ = 10
The hypotheses can be outlined as:
The critical value is t₃₆+₃₆-₂,₀.₀₁ = t₇₀,₀.₀₁
thus,
t₃₆+₃₆-₂,₀.₀₁ = t₇₀,₀.₀ = 2.381
This implies a range of -2.381 to 2.381
Thus, we can support the professor's assertion.