Good evening ,
Answer :
A.
x = ±20
Step-by-step explanation:
x² − 1 = 399 ⇔ x²= 400 ⇌ x² = 20² ⇌ x² − 20² = 0 ⇌ (x-20)(x+20) = 0
⇌ x = 20 or x = -20.
:)
Answer:
The correct option is;

Step-by-step explanation:
Here we have the formula for the confidence interval of the difference of two means where the population standard deviation is unknown based on the sample mean and sample standard deviation is given as follows;

Where:
= Mean of the first sample = 22 grams
= Mean of the second sample = 18 grams
s₁ = Sample standard deviation of the first sample = 3.2 grams
s₂ = Sample standard deviation of the second sample = 2.1 grams
n₁ = Sample size of the first sample = 100
n₂ = Sample size of the second sample = 100
= t value obtained from tables at 99% confidence level and 100 degrees of freedom = 2.626 = 2.63
Therefore, plugging in the values, we have;

Therefore, the correct option is
.
Answer:
0ft and 60ft
Step-by-step explanation:
Given
The attached function
Required
Determine the valid values of the domain of the function
To do this, we simply consider the starting point and the end point of the trajectory on the x-axis (i.e. the horizontal distance).
From the attached graph, the horizontal distance starts from 0 and ends at 180.
This implies that the domain is: 
From the options, the values that fall in this bracket are 0ft and 60ft
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