Solve each equation:

Suppose that 80% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou

Gwar [14]

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

7 0

2 years ago

Sidh is making fruit salad. His recipe calls for 3/4 pound of watermelon. He wants

Dvinal [7]

Answer:

3/8

Step-by-step explanation:

3/4 x 1/2 = 3/8

6 0

3 years ago

a 9 pound bag of sugar is being split into containers that hold 2/3 of a pound. how many containers of sugar will the 9 pound ba

LUCKY_DIMON [66]

Namely, how many times does 2/3 go into 9?

well

$\bf 9 \div \frac{2}{3}\implies \cfrac{9}{\frac{2}{3}}\implies \cfrac{\quad \frac{9}{1}\quad }{\frac{2}{3}}\implies \cfrac{9}{1}\cdot \cfrac{3}{2}\implies \cfrac{27}{2}
\\\\\\
\textit{2 goes \underline{13 times} into 27, with a \underline{remainder of 1}}\implies 13\frac{1}{2}$

3 0

3 years ago

Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m

Dovator [93]

Option a: $\frac{m^{5} }{162n}$ is the equivalent expression.

Explanation:

The expression is $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ where $m\neq 0, n\neq 0$

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

$\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }$

Cancelling the like terms, we have,

$\frac{3^{-3}m^{5} n^{-1}}{6 }$

This equation can also be written as,

$\frac{m^{5}}{3^{3}6 n^{1} }$

Multiplying the terms in denominator, we have,

$\frac{m^{5} }{162n}$

Thus, the expression which is equivalent to $\frac{(3m^{-2} n)^{-3}}{6mn^{-2} }$ is $\frac{m^{5} }{162n}$

Hence, Option a is the correct answer.

8 0

3 years ago

Read 2 more answers

Every prime number greater than 10 has a digit in the ones place that is include in wich set of numbers A- 1,3,7,9 B-1,3,5,9 D-1

marta [7]

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' .

4 0

3 years ago