Answer:
For first lamp ; The resultant probability is 0.703
For both lamps; The resultant probability is 0.3614
Step-by-step explanation:
Let X be the lifetime hours of two bulbs
X∼exp(1/1400)
f(x)=1/1400e−1/1400x
P(X<x)=1−e−1/1400x
X∼exp(1/1400)
f(x)=1/1400 e−1/1400x
P(X<x)=1−e−1/1400x
The probability that both of the lamp bulbs fail within 1700 hours is calculated below,
P(X≤1700)=1−e−1/1400×1700
=1−e−1.21=0.703
The resultant probability is 0.703
Let Y be a lifetime of another lamp two bulbs
Then the Z = X + Y will follow gamma distribution that is,
X+Y=Z∼gamma(2,1/1400)
2λZ∼
X+Y=Z∼gamma(2,1/1400)
2λZ∼χ2α2
The probability that both of the lamp bulbs fail within a total of 1700 hours is calculated below,
P(Z≤1700)=P(1/700Z≤1.67)=
P(χ24≤1.67)=0.3614
The resultant probability is 0.3614
Answer:
Here is your solution-
x³ times x⁷
means, x³×x⁷
Here you just have to add the powers as aⁿ×aᵐ=aⁿ⁺ᵐ
x³⁺⁷
x¹⁰
It’s green:b purple:d blue:c red:a because
Answer:
There are no solutions.
Step-by-step explanation:
4x+5=6x+18-20-2x (Distributive Property)
4x+2x-6x=18-20+5 (Combine Like Terms)
0=3 (Simplify)
That doesn't make any sense, so there are no solutions.
