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algol13
2 years ago
15

Edg vector operations, any help appreciated!

Mathematics
1 answer:
viktelen [127]2 years ago
8 0

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: Add \:\: -6 \hat i - 6\hat j \:\:with \:\; Vector \:\; c

____________________________________

\large \tt Solution  \: :

Vector d can be represented as :

\qquad \tt \rightarrow \:  - 2 \hat i - 2 \hat j

Vector c can be represented as :

\qquad \tt \rightarrow \:  4 \hat i + 4\hat j

we have to create vector d from vector c

So, let's assume a vector x, such that sum of vector x and vector c equals to vector d

\qquad \tt \rightarrow \: x + ( 4 \hat i + 4 \hat j) =  - 2 \hat i  - 2 \hat j

\qquad \tt \rightarrow \: x  = -  ( 4 \hat i + 4 \hat j)   - 2 \hat i  - 2 \hat j

\qquad \tt \rightarrow \: x  = (-   4 \hat i  - 2 \hat  i)  + (  - 4 \hat j  - 2 \hat j)

\qquad \tt \rightarrow \: x  = - 6 \hat  i    -6 \hat j

Henceforth, in order to get vector d, we need to add (-6i - 6j) in vector c

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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Yakvenalex [24]

Answer:

a) 98.01%

b) 13.53\%

c) 27.06%

Step-by-step explanation:

Since a car has 10 square feet of plastic panel, the expected value (mean) for a car to have one flaw is 10*0.02 = 0.2  

If we call P(k) the probability that a car has k flaws then, as P follows a Poisson distribution with mean 0.2,

P(k)= \frac{0.2^ke^{-0.2}}{k!}

a)

In this case, we are looking for P(0)

P(0)= \frac{0.2^0e^{-0.2}}{0!}=e^{-0.2}=0.9801=98.01\%

So, the probability that a car has no flaws is 98.01%

b)

Ten cars have 100 square feet of plastic panel, so now the mean is 100*0.02 = 2 flaws every ten cars.

Now P(k) is the probability that 10 cars have k flaws and  

P(k)= \frac{2^ke^{-2}}{k!}

and  

P(0)= \frac{2^0e^{-2}}{0!}=0.1353=13.53\%

And the probability that 10 cars have no flaws is 13.53%

c)

Here, we are looking for P(1) with P defined as in b)

P(1)= \frac{2^1e^{-2}}{1!}=2e^{-2}=0.2706=27.06\%

Hence, the probability that at most one car has no flaws is 27.06%

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