Answer:
Domain = (
-∞,∞), {x|x ∈ R}
Range (-∞,2], {y|y ≤ 2}
Vertex (h,k) = (6,2)
Step-by-step explanation:
(Domain / Range) The absolute value expression has a V shape. The range of a negative absolute value expression starts at its vertex and extends to negative infinity.
(Vertex) To find the x coordinate of the vertex, set the inside of the absolute value
x − 6 equal to 0 . In this case, x − 6 = 0 .
x−6=0
Add 6 to both sides of the equation.
x=6
Replace the variable x with 6 in the expression.
y=−1/3⋅|(6)−6|+2
Simplify−1/3⋅|(6)−6|+2.
y=2
The absolute value vertex is ( 6 , 2 ) .
(6,2)
Hope this helps
Answer:
Step-by-step explanation:
This is not a question we need to at least know it’s equation.
Steps for multiplying the given binomial and trinomial 
are:
Step 1: 
Step 2: 
Step 3: 
Answer: 20
Step-by-step explanation:
Given
The cost of an individual ticket is $25
The cost of a couple's ticket is $40
The total sale is $2500
total ticket sold is 70
Suppose there are x individuals and y couples
So, they sold 20 tickets of the individual.
The standard deviation is 9.27. The typical heart rate for the data set varies from the mean by an average of 9.27 beats per minute.
<h3>How to determine the standard deviation of the data set?</h3>
The dataset is given as:
Heart Rate Frequency
60 1
65 3
70 4
75 12
80 8
85 15
90 9
95 5
100 3
Calculate the mean using
Mean = Sum/Count
So, we have
Mean = (60 * 1 + 65 * 3 + 70 * 4 + 75 * 12 + 80 * 8 + 85 * 15 + 90 * 9 + 95 * 5 + 100 * 3)/(1 + 3 + 4 + 12 + 8 + 15 + 9 + 5 + 3)
Evaluate
Mean = 82.25
The standard deviation is
So, we have:
SD = √[1 * (60 - 82.25)^2 + 3 * (65 - 82.25)^2 + 4 * (70 - 82.25)^2 + 12 * (75 - 82.25)^2 + 8 * (80 - 82.25)^2 + 15 * (85 - 82.25)^2 + 9 * (90 - 82.25)^2 + 5 * (95 - 82.25)^2 + 3 * (100 - 82.25)^2)]/[(1 + 3 + 4 + 12 + 8 + 15 + 9 + 5 + 3 - 1)]
This gives
SD = √85.9533898305
Evaluate
SD = 9.27
Hence. the standard deviation is 9.27. The typical heart rate for the data set varies from the mean by an average of 9.27 beats per minute.
Read more about standard deviation at:
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