Answer:
yes
Step-by-step explanation:
Substitute x = 2 and y = 5 into the left side of the equations and if equal to the right side then they are a solution.
x + 2y = 2 + 2(5) = 2 + 10 = 12 ← True
3x + 6y = 3(2) + 6(5) = 6 + 30 = 36 ← True
Thus x = 2, y = 5 is a solution to both equations
(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:
For the integral, substitute <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:
which reduces to
<em>c</em> / 2 (0 + 1) = 1 → <em>c</em> = 2
(b) Find the probability P(1 < <em>X </em>< 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):
Answer:
(a) x = -2y
(c) 3x - 2y = 0
Step-by-step explanation:
You can tell if an equation is a direct variation equation if it can be written in the format y = kx.
Note that there is no addition and subtraction in this equation.
Let's put these equations in the form y = kx.
(a) x = -2y
- y = x/-2 → y = -1/2x
- This is equivalent to multiplying x by -1/2, so this is an example of direct variation.
(b) x + 2y = 12
- 2y = 12 - x
- y = 6 - 1/2x
- This is not in the form y = kx since we are adding 6 to -1/2x. Therefore, this is <u>NOT</u> an example of direct variation.
(c) 3x - 2y = 0
- -2y = -3x
- y = 3/2x
- This follows the format of y = kx, so it is an example of direct variation.
(d) 5x² + y = 0
- y = -5x²
- This is not in the form of y = kx, so it is <u>NOT</u> an example of direct variation.
(e) y = 0.3x + 1.6
- 1.6 is being added to 0.3x, so it is <u>NOT</u> an example of direct variation.
(f) y - 2 = x
- y = x + 2
- 2 is being added to x, so it is <u>NOT</u> an example of direct variation.
The following equations are examples of direct variation:
Answer:
The mom would offer her 150% of 350
=> 150/100 * 350
=> 3/2 * 350
=> 1050/2
=> 525
Therefore, the mom would offer her 525 dollars
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