B I guess. since they both have potential to collapse
Answer:
0.264 ; 0.079
Explanation:
Given that:
Sample size, n = 100
Probability of being active, p = 1% = 1/100 = 0.01
Using the binomial probability relation :
P(x =x) = nCx * p^x * (1 - p)^(n - x)
Probability that more than 1 user will be active
P(x > 1) = 1 - [p(x=0) + p(x = 1)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x > 1) = 1 - [0.366 + 0.370]
P(x > 1) = 0.264
2.)
Probability that more than 2 user will be active
P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185
P(x > 1) = 1 - [0.366 + 0.370 + 0.185]
P(x > 1) = 0.079
Answer:
Use a screw or corkscrew. Screw it into the eraser, then pull.
Answer:
- resistance (ohms) will stay the same
- amps will double
Explanation:
Assuming linear circuit elements being operated within their ratings, ...
- resistance (ohms) will stay the same
- amps will double
For constant resistance, current is proportional to voltage. For a circuit in which resistance is not affected by voltage or temperature, the resistance will stay the same when voltage increases. Then the current will be proportional to the voltage.
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<em>Additional comments</em>
In some circuits, resistance is a function of temperature. An incandescent light, for example, increases its resistance as its temperature goes up. When voltage is increased, resistance is increased according to the increased power dissipation. Current increases, but not proportionally to voltage.
Some other devices also exhibit this effect, but their resistance is designed to increase to a high enough level that overall circuit current actually decreases when voltage exceeds a certain threshold.
