All the cars in the lot of a particular dealership have either a manual transmission or an automatic transmission, and all have either two doors or four doors. in this lot, 60% of the cars have a manual transmission and 70% of the cars have four doors. if 90% of the cars in the lot have either a manual transmission or four doors or both, then what percentage of the automatic transmission cars have four doors?

30%

Step-by-step explanation:

60% of the cars have manual transmission. This means P(Manual) = 60%

70% of the cars have four doors. This means P(Four Doors) = 70%

90% of the cars have either manual transmission or four doors or both. This means: P(Manual OR Four doors) = 90%

The formula of probability in case of OR(Union) of two events A and B is:

P(A or B) = P(A) + P(B) - P(A and B)

Re-writing the formula for our case, we get:

P(Manual or Four Doors) = P(Manual) + P(Four doors) - P(Both Manual and Four Doors)

Using the given values, we get:

90% = 60% + 70% - P(Both Manual and Four Doors)

90% = 130% - P(Both Manual and Four Doors)

P(Both Manual and Four Doors) = 130% - 90%

P(Both Manual and Four Doors) = 40%

This means, 40% vehicles have both Manual Transmission and Four Doors.

Since, in total 70% of the cars have 4 doors, out of these 40% are manual. This means, the remaining 30% of the cars with 4 doors have automatic transmission.

Hence, 30% of the automatic transmission cars have four doors

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5