Answer:
x=-3
Step-by-step explanation:
(3x-15)/2 = 4x
Multiply each side by 2
(3x-15)/2 *2= 4x*2
3x-15 = 8x
Subtract 3x from each side
3x-15-3x = 8x-3x
-15 = 5x
Divide each side by 5
-15/5 = 5x/5
-3 =x
Answer:
The multiplicative inverse of -0.7 is -1/0.7
hope it helps
<span>2/15 if drawn without replacement.
1/9 if drawn with replacement.
Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are:
1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5
Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15
If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are:
1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6
And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>
Answer:
a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90
a= μ-3.16*σ , b= μ+3.16*σ
b) P(Y≥ μ+3*σ ) ≥ 0.90
b= μ+3*σ
Step-by-step explanation:
from Chebyshev's inequality for Y
P(| Y - μ|≤ k*σ ) ≥ 1-1/k²
where
Y = the number of fish that need be caught to obtain at least one of each type
μ = expected value of Y
σ = standard deviation of Y
P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean
k= parameter
thus for
P(| Y - μ|≤ k*σ ) ≥ 1-1/k²
P{a≤Y≤b} ≥ 0.90 → 1-1/k² = 0.90 → k = 3.16
then
P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90
using one-sided Chebyshev inequality (Cantelli's inequality)
P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)
P{Y≥b} ≥ 0.90 → 1- σ²/(σ²+λ²)= 1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ
then for
P(Y≥ μ+3*σ ) ≥ 0.90
