I'm not 100% sure I'm just trying to help. I've seen this question before and I believe the answer was <span>362,880, but that seems like an a full lot to me personally, but like I said just trying to help. :)</span>
24.75= 2.25y - 22.95
+22.95 +22.95
47.7 = 2.25y (division here)
2.25 2.25
Y=21.2
<h3>
Answer: g(x) = (-2/3)*x^2</h3>
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Work Shown:
f(x) = x^2
g(x) = a*f(x) for some constant 'a' since g(x) is a scaled version of f(x).
The value of 'a' vertically stretches f(x) upward if a > 0
If 'a' is negative, then we have a reflection going on as shown in the diagram.
We want (x,y) = (3,-6) to be on the graph of g(x). This means g(3) = -6
If we plugged x = 3 into f(x), we get
f(x) = x^2
f(3) = 3^2
f(3) = 9
So,
g(x) = a*f(x)
g(3) = a*f(3) ... replace x with 3
g(3) = a*9 ... replace f(3) with 9 since f(3) = 9
-6 = a*9 ... replace g(3) with -6 since g(-3) = -6
9a = -6
a = -6/9
a = -2/3
Therefore, this means
g(x) = a*f(x)
g(x) = (-2/3)*f(x)
g(x) = (-2/3)*x^2
Answer:
the probability that a box weighs exactly 31 ounces is 0
Step-by-step explanation:
since the box has a continuous distribution (therefore there are infinite real numbers between 31.8 and 32.6 ounces) , the probability of a exact value out of a infinite quantity is 0 . Mathematically , since the probability for a random variable X to be between values of 'a' and 'b' for a continuous probability density function f(x) is
P(a≤X≤b)=∫ₐᵇ f(x) dx
for b=a, then
P(X=a)= ∫ₐᵃ f(x) dx = 0
Nevertheless , since the value of 31 is outside the range of 31.8 and 32.6 ounces, the probability of a value outside the continuous distribution is also 0
