Answer: -6
Step-by-step explanation: It changed -6 each round because -6x3=-18.\
Hope this helped!
Let <em>a</em> and <em>b</em> be the zeroes of <em>x</em>² + <em>kx</em> + 12 such that |<em>a</em> - <em>b</em>| = 1.
By the factor theorem, we can write the quadratic in terms of its zeroes as
<em>x</em>² + <em>kx</em> + 12 = (<em>x</em> - <em>a</em>) (<em>x</em> - <em>b</em>)
Expand the right side and equate the coefficients:
<em>x</em>² + <em>kx</em> + 12 = <em>x</em>² - (<em>a</em> + <em>b</em>) <em>x</em> + <em>ab</em>
Then
<em>a</em> + <em>b</em> = -<em>k</em>
<em>ab</em> = 12
The condition that |<em>a</em> - <em>b</em>| = 1 has two cases, so without loss of generality assume <em>a</em> > <em>b</em>, so that |<em>a</em> - <em>b</em>| = <em>a</em> - <em>b</em>.
Then if <em>a</em> - <em>b</em> = 1, we get <em>b</em> = <em>a</em> - 1. Substitute this into the equations above and solve for <em>k</em> :
<em>a</em> + (<em>a</em> - 1) = -<em>k</em> → 2<em>a</em> = 1 - <em>k</em> → <em>a</em> = (1 - <em>k</em>)/2
<em>a</em> (<em>a</em> - 1) = 12 → (1 - <em>k</em>)/2 • ((1 - <em>k</em>)/2 - 1) = 12
→ (1 - <em>k</em>)²/4 - (1 - <em>k</em>)/2 = 12
→ (1 - <em>k</em>)² - 2 (1 - <em>k</em>) = 48
→ (1 - 2<em>k</em> + <em>k</em>²) - 2 (1 - <em>k</em>) = 48
→ <em>k</em>² - 1 = 48
→ <em>k</em>² = 49
→ <em>k</em> = ± √(49) = ±7
<span>In a jar of ten beads; since 3 are blue; probability of picking a blue ball, B, = p(b) = 3/10. And P (of not picking a blue ball) ; p(b') = 7/10.
Since it occurs with replacement, probabilities doesn't change
Probaility of picking k blue balls from on n attempts is given by P_n(k)
P_n(k) = (n, k) p^(k) q^(n -k) where p and q are b and b' respectively.
P_5(2) = (5 , 2) (0.3)^(2) (0.7)^(5 - 2)
P_5(2) = 5C2 (0.3)^(2) (0.7)^(3) = 0.3087</span>
