Q=56??? Sorry if it’s wrong lol
Answer:
g(9) = 6
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Step-by-step explanation:
<u>Step 1: Define</u>
g(x) = -x² + 10x - 3
g(9) is x = 9
<u>Step 2: Evaluate</u>
- Substitute in <em>x</em>: g(9) = -(9)² + 10(9) - 3
- Exponents: g(9) = -(81) + 10(9) - 3
- Multiply: g(9) = -81 + 90 - 3
- Add: g(9) = 9 - 3
- Subtract: g(9) = 6
Answer:
0.1384
Step-by-step explanation:
Using binomial probability:
P = nCr p^r q^(n-r)
where n is the number of trials,
r is the number of successes,
p is the probability of success,
and q is the probability of failure (1-p).
Given n = 25, r = 4, p = 0.10, and q = 0.90:
P = ₂₅C₄ (0.10)⁴ (0.90)²⁵⁻⁴
P = 12650 (0.10)⁴ (0.90)²¹
P = 0.1384
The equation of the line through (0, 1) and (<em>c</em>, 0) is
<em>y</em> - 0 = (0 - 1)/(<em>c</em> - 0) (<em>x</em> - <em>c</em>) ==> <em>y</em> = 1 - <em>x</em>/<em>c</em>
Let <em>L</em> denote the given lamina,
<em>L</em> = {(<em>x</em>, <em>y</em>) : 0 ≤ <em>x</em> ≤ <em>c</em> and 0 ≤ <em>y</em> ≤ 1 - <em>x</em>/<em>c</em>}
Then the center of mass of <em>L</em> is the point 
with coordinates given by
where 
is the first moment of <em>L</em> about the <em>x</em>-axis, 
is the first moment about the <em>y</em>-axis, and <em>m</em> is the mass of <em>L</em>. We only care about the <em>y</em>-coordinate, of course.
Let <em>ρ</em> be the mass density of <em>L</em>. Then <em>L</em> has a mass of
Now we compute the first moment about the <em>y</em>-axis:
Then
but this clearly isn't independent of <em>c</em> ...
Maybe the <em>x</em>-coordinate was intended? Because we would have had
and we get
