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3 years ago

Point B has rectangular coordinates (-5, 12)

Mathematics

1 answer:

Irina-Kira [14]3 years ago

7 0

Answer:

(13, -67.4°)

Step-by-step explanation:

In the general case, for a point (x, y), the same point written in polar coordinates is (r, θ), such that:

r = √(x^2 + y^2)

θ = Atg(y/x)

Then if we have the point with coordinates (-5, 12)

The polar coordinates will be:

r = √((-5)^2 + 12^2) = 13

θ = Atg(12/-5) = -67.4°

Then the polar coordinates are:

(13, -67.4°)

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4 years ago

In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien

diamong [38]

Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = $\frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}$

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) = $\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)}$ = $\frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)}$ = 0.1138

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4 years ago

Od 2 ( 2 7 1-3 1 3 1-2 6:1-6 b = d = IDONE

Lena [83]

Answer:

$a = \boxed{1},\: b=\boxed{0}\\\\c=\boxed{0},\: d=\boxed{1}$

Step-by-step explanation:

$\begin{bmatrix} 2 & 7\\\\1 & 3\end{bmatrix}\:\begin{bmatrix} -3 & 7\\\\1 & -2\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}$

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$\begin{bmatrix} 2(-3)+7(1) & 2(7)+7(-2)\\\\1(-3)+3(1) & 1(7)+3(-2)\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}$

$\rightarrow\begin{bmatrix} -6+7 & 14-14\\\\-3+3 & 7-6\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}$

$\rightarrow\begin{bmatrix} 1 & 0\\\\0 & 1\end{bmatrix}=\begin{bmatrix} a & b\\\\c & d\end{bmatrix}$

Comparing the corresponding elements of both the matrices on both sides, we find:

$a = \boxed{1},\: b=\boxed{0},\:c=\boxed{0},\: d=\boxed{1}$

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