Answer:
P=0.147
Step-by-step explanation:
As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2
We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.
We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.
So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7
P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02
P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007
P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12
P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147
Answer:
3/8
Step-by-step explanation:
3/4 x 1/2 = 3/8
Option a:
is the equivalent expression.
Explanation:
The expression is
where 
Let us simplify the expression, to determine which expression is equivalent from the four options.
Multiplying the powers, we get,

Cancelling the like terms, we have,

This equation can also be written as,

Multiplying the terms in denominator, we have,

Thus, the expression which is equivalent to
is 
Hence, Option a is the correct answer.
That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:
-- 0:
A number greater than 10 with a 0 in the units place is a multiple of
either 5 or 10, so it's not a prime number.
-- 1:
A number greater than 10 with a 1 in the units place could be
a prime (11, 31 etc.) but it doesn't have to be (21, 51).
-- 2:
A number greater than 10 with a 2 in the units place has 2 as a factor
(it's an even number), so it's not a prime number.
-- 3:
A number greater than 10 with a 3 in the units place could be
a prime (13, 23 etc.) but it doesn't have to be (33, 63) .
-- 4:
A number greater than 10 with a 4 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 5:
A number greater than 10 with a 5 in the units place is a multiple
of either 5 or 10, so it's not a prime number.
-- 6:
A number greater than 10 with a 6 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 7:
A number greater than 10 with a 7 in the units place could be
a prime (17, 37 etc.) but it doesn't have to be (27, 57) .
-- 8:
A number greater than 10 with a 8 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 9:
A number greater than 10 with a 9 in the units place could be
a prime (19, 29 etc.) but it doesn't have to be (39, 69) .
So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.
It CAN't have a 0, 2, 4, 5, 6, or 8 .
The only choice that includes all of the possibilities is 'A' .