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3 years ago

3cis(5pi/4) in polar form

1 answer:

8 0

Rectangular form:
z = -2.1213203-2.1213203i

Angle notation (phasor):
z = 3 ∠ -135°

Polar form:
z = 3 × (cos (-135°) + i sin (-135°))

Exponential form:
z = 3 × ei (-0.75) = 3 × ei (-3π/4)

Polar coordinates:
r = |z| = 3 ... magnitude (modulus, absolute value)
θ = arg z = -2.3561945 rad = -135° = -0.75π = -3π/4 rad ... angle (argument or phase)

Cartesian coordinates:
Cartesian form of imaginary number: z = -2.1213203-2.1213203i
Real part: x = Re z = -2.121
Imaginary part: y = Im z = -2.12132034

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What is the 14 called in 14^2 -exponent, radical sign or base

gayaneshka [121]

Answer:

Base

Step-by-step explanation:

An exponent is a number/letter written to the top-right of a mathematical expression called the base. It means that the base will raised to a certain power.

The radical sign is a symbol for the square root of a number.

6 0

3 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0

3 years ago

The range of a data set is 32. If the largest number in the set is 50, what is the smallest number?

TEA [102]

Answer:

Step-by-step explanation:

50 - 32 = 18

7 0

3 years ago

Need help please help

azamat

Answer:

Hope you get it right! :)

Step-by-step explanation:

7 0

3 years ago

Consider the quadratic equation x? = 4x - 5. How many solutions does the equation have? A The equation has one real solution. Th

anzhelika [568]

Answer:

{-1, 5} (two real solutions).

Step-by-step explanation:

On the left side we want x^2, not x?

We need to rewrite x? = 4x - 5 in standard form, that is, as a quadratic with x terms in decreasing powers of x:

x^2 - 4x + 5 = 0

This factors easily to (x - 5)(x + 1) = 0.

We let each factor equal zero (0) separately and solve for x:

{-1, 5} (two real solutions).

4 0

3 years ago