Step 1: Recall properties of exponential distribution.
For an exponential distribution with parameter \( \lambda = 2 \), the expected value is:
\[
E(X) = \frac{1}{\lambda} = \frac{1}{2}
\]
Step 2: Use the survival function.
The probability that the lifetime exceeds a value \( t \) is:
\[
P(X > t) = e^{-\lambda t}
\]
Step 3: Substitute the expected lifetime.
\[
P\left(X > \frac{1}{2}\right) = e^{-2 \times \frac{1}{2}} = e^{-1}
\]
Step 4: Numerical approximation.
\[
e^{-1} \approx 0.3679
\]
Step 5: Round to two decimal places.
\[
P(X > E(X)) \approx 0.39
\]
% Final Answer
Final Answer: \[ \boxed{0.39} \]
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int x = 126, y = 105;
do {
if (x > y)
x = x - y;
else
y = y - x;
} while (x != y);
printf("%d", x);
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| Multiplicand (\( M \)) | Multiplier (\( Q \)) |
|---|---|
| 1100 1101 1110 1101 | 1010 0100 1010 1010 |
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Consider the following C program
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