If the probability of Bob passing a driving test is p and the probability he fails is 6p^2 what is the value of p?

We are given:

Probability of passing = p

Probability of Failing = 6p²

Finding the value of 'p':

We know that the sum of the probabilities of an event occurring and not occurring is 1

So, Probability of passing + Probability of Failing = 1

p + 6p² = 1

6p² + p - 1 = 0

We can see that we now have a quadratic equation

Splitting the middle term:

6p² + 3p - 2p - 1 = 0

3p(2p + 1) - (2p + 1) = 0

(3p - 1)(2p + 1) = 0                      [factoring out 2p + 1]

Now, we can either divide both sides by (3p-1) or by (2p + 1) and both the cases will give us different answers.

Dividing both sides by (3p - 1):

2p + 1 = 0/(3p-1)

2p + 1 = 0

p = -1/2

Dividing both sides by (2p+1):

(3p - 1) = 0 / (2p + 1)

3p - 1 = 0

p = 1/3

Now, we have 2 values of p: -1/2 and 1/3

Using the value: -1/2 will mean that the probabilities are negative

but we know that probabilities cannot be negative.

So, -1/2 is not a valid value for p

Therefore, the value of p is 1/3

all but the second one

step-by-step explanation:

cuz that one goes over 180

mark brainliest

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v