A consumer organization estimates that over a one-year period: 17% of cars will need to be repaired one time
7% of cars will need to be repaired two times
4% of cars will need to be repaired three or more times

What is the probability that a car chosen at random will need:
no repairs?

no more than one repair? (think of it as one repair or fewer)

some repairs? (think of it as needing one, two, or three repairs)

If you own two cars, what is the probability that:
neither will need to be repaired?

both will need to be repaired?

at least one car will need to be repaired?

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day

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

after going and makeing graphs i would go with a,

there is nothing that is matches the graph. if there was i would show you. but therefore my answer is a.

hope i

-lavira: )