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3 years ago

BRAINLIESTT ASAP! PLEASE HELP ME :)

Mathematics

2 answers:

Rudiy273 years ago

8 0

<h3>Answer: D) none of the equations are identities</h3>

====================================

Work Shown:

I'm going to use x in place of theta.

let

f(x) = sin(x+2pi/4)

g(x) = sqrt(2)*sin(x)*cos(x)

Plug in x = 0

f(x) = sin(x+2pi/4)

f(0) = sin(0+2pi/4)

f(0) = sin(2pi/4)

f(0) = sin(pi/2)

f(0) = 1

and

g(x) = sqrt(2)*sin(x)*cos(x)

g(0) = sqrt(2)*sin(0)*cos(0)

g(0) = sqrt(2)*0*1

g(0) = 0

We see that f(0) = g(0) is not a true equation

Therefore, f(x) = g(x) is not true for all values of x, making it not be an identity. We can use a graph to compare f(x) and g(x) to see that they do not line up.

--------------------------------

Let

h(x) = cos(x-pi/3)

i(x) = (3/sqrt(2))*(cos(x) - sin(x))

Plug in x = 0

h(x) = cos(x-pi/3)

h(0) = cos(0-pi/3)

h(0) = 0.5

and

i(x) = (3/sqrt(2))*(cos(x) - sin(x))

i(0) = (3/sqrt(2))*(cos(0) - sin(0))

i(0) = 2.12

Like with equation 1, we do not have an identity since h(0) does not equal i(0). We only need one counter example since an identity states that h(x) = i(x) for all values of x in the domain.

Makovka662 [10]3 years ago

7 0

Answer:

Step-by-step explanation:

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The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

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Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

2/21 x 3 <br> Multiply. Write your answer as a fraction in the simplest form.

Mnenie [13.5K]

Answer:

2/7

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Suppose that the times required for a cable company to fix cable problems in its customers’ homes are uniformly distributed betw

vampirchik [111]

Answer:

a) 60% probability that a randomly selected cable repair visit will take at least 50 minutes

b) 60% probability that a randomly selected cable repair visit will take at most 55 minutes.

c) The expected length of the repair visit is 52.5 minutes.

d) The standard deviation is 7.22 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

For this problem, we have that:

Uniformly distributed between 40 and 65 minutes, so $a = 40, b = 65$

(a) What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

Either it takes less than 50 minutes, or it takes at least 50 minutes. The sum of the probabilities of these events is decimal 1. So

$P(X < 50) + P(X \geq 50) = 1$

The intervals can be open or closed, so

$P(X < 50) = P(X \leq 50)$

We have that

$P(X \geq 50) = 1 - P(X < 50)$

$P(X < x) = \frac{50 - 40}{65 - 40} = 0.4$

$P(X \geq 50) = 1 - P(X < 50) = 1 - 0.4 = 0.6$

60% probability that a randomly selected cable repair visit will take at least 50 minutes

(b) What is the probability that a randomly selected cable repair visit will take at most 55 minutes?

This is $P(X \leq 55)$

$P(X \leq 55) = \frac{55 - 40}{65 - 40} = 0.6$

60% probability that a randomly selected cable repair visit will take at most 55 minutes.

(c) What is the expected length of the repair visit?

The mean of the uniform distribution is:

$M = \frac{a+b}{2}$

$M = \frac{40+65}{2} = 52.5$

The expected length of the repair visit is 52.5 minutes.

(d) What is the standard deviation?

The standard deviation of the uniform distribution is

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

$S = \sqrt{\frac{(65-40)^{2}}{12}}$

$S = 7.22$

The standard deviation is 7.22 minutes.

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3 years ago

A parallelogram piece of paper 40cm by 25cm is cut into small parallelogram 8cm by5cm . How many pieces can be obtained. Step by

Bumek [7]

Step 1- multiply 40 by 25 to get 1000
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step3- divide 1000 by 4 to get 250
answer. 250 pieces

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3 years ago

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X=4.8
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3 years ago