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Crazy boy [7]
3 years ago
13

Components of a certain type are shipped to a supplier in batches of ten. suppose that 50% of all such batches con- tain no defe

ctive components, 30% contain one defective component, and 20% contain two defective components. two compone
Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0
Solution
Let 0 be the event that the batch has 0 defectives, 1 be the event the batch has 1 defective, and 2 be
the event the batch has 2 defectives. Let 0 be the event that neither selected component is defective.
(0
) = 0.5,(1
) = 0.3, (2
) = 0.2
The event 0 can happen in three different ways: (i) Our batch of 10 is perfect, and we get no defectives in
our sample of two; (ii) Our batch of 10 has 1 defective, but our sample of two misses them; (iii) Our batch
has 2 defective, but our sample misses them.
For (i), the probability is (0.5)(1).
For (ii), the probability that our batch has 1 defective is 0.3. Given that it has 1 defective, the probability
that our sample misses it is (
9
2
)
(
10
2
)
, which is 8
10
. So the probability of (ii) is (0.3) (
8
10).
For (iii), the probability our batch has 2 defective is 0.2. Given that it has 2 defective, the probability that
our sample misses both is (
8
2
)
(
10
2
)
, which is 56
90
. So the probability of (iii) is (0.2) (
56
90). We have therefore found
that
(0
) = (0.5)(1) + (0.3) (
8
10) + (0.2) (
56
90).
We use the general conditional probability formula:
(0
|0
) =
(0 ∩ 0
)
(0
)
=
(0.5)(1)
(0.5)(1) + (0.3) (
8
10) + (0.2) (
56
90)
= 0.5784.
(1
|0
) =
(1∩0
)
(0
)
=
(0.3)(
8
10)
(0.5)(1)+(0.3)(
8
10)+(0.2)(
56
90)
= 0.2776.
(2
|0
) =
(2∩0
)
(0
)
=
(0.2)(
56
90)
(0.5)(1)+(0.3)(
8
10)+(0.2)(
56
90)
= 0.1440.
Answer: 0.5784; 0.2776; 0.1440.
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7 0
3 years ago
CAN SOMEONE MULTIPLY THIS PLEASE, ITS URGENT
____ [38]

Let's factorise it :

\: {\qquad  \dashrightarrow \sf   ( {x}^{3}   -  5)(x + 3)  }

\: {\qquad  \dashrightarrow \sf    {x}^{3} (x + 3) + [-5(x + 3)]  }

Using Distributive property we get :

\: {\qquad  \dashrightarrow \sf    {x}^{3} +  {3x}^{3}   + ( - 5x - 15)  }

\: {\qquad  \dashrightarrow \sf    {x}^{3} +  {3x}^{3}   - 5x - 15 }

\: {\qquad  \dashrightarrow \sf    4{x}^{3}  - 5x - 15 }

⠀

Therefore,

\: {\qquad  \dashrightarrow \sf   ( {x}^{3}   -  5)(x + 3)   =4{x}^{3}  - 5x - 15}

8 0
2 years ago
The school paper charges $1.50 for an ad, plus 25 cents per word. Carl wants to spend only $12.00 on the ad.
statuscvo [17]

Answer:

42 words

Step-by-step explanation:

To solve just solve 0.25x +1.50 = 12.00 for x

7 0
3 years ago
Read 2 more answers
alvin and simon raced a distance of 2.9 x 10^2 kilometers. they need to race 7 times that distance in kilometer. how far must th
dimaraw [331]

The distance in kilometers in scientific notation 2.03 \times 10^3 km

<h3>Scale factors</h3>

Scale factors are values or constants multiplied with a variable

If Alvin and Simon raced a distance of 2.9 \times 10^2 kilometers and they need to race 7 times that distance in a kilometer, the distance traveled will be:

  • Distance travelled = 7(2.9 \times 10^2)
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Hence the distance in kilometers in scientific notation 2.03 \times 10^3 km

Learn more on distance here: brainly.com/question/23848540

8 0
2 years ago
HURRY ITS URGENT!!!!
MArishka [77]
It has to be either A or B
5 0
2 years ago
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