Please answer !!!!!!!!!!!!

Which statements are true? Select all that apply.

AveGali [126]

cosθ = cotθ/cscθ is a true statement. The answer is option B

<h3>How to determine which of the trigonometric statements are true?</h3>

Trigonometry is a branch of mathematics dealing with the relationship between the ratios of the sides of a right-angled triangle with its angles

A. tan²θ = 1 - sec²θ

tan²θ = 1 - sec²θ

tan²θ = 1 - 1/cos²θ (Note: sec²θ = 1/cos²θ)

tan²θ = (cos²θ- 1)/cos²θ

tan²θ = -sin²θ/cos²θ (Note: cos²θ- 1 = -sin²θ)

tan²θ = -tan²θ

This statement is not true

B. cosθ = cotθ/cscθ

cosθ = cotθ/cscθ

cosθ = (1/tanθ) / (1/sinθ)

cosθ = (cosθ/sinθ).sinθ

cosθ = cosθ

This statement is true

C. 1/sec²θ = sin²θ + 1

1/sec²θ = 1/(1/cos²θ)

1/sec²θ = cos²θ

1/sec²θ = 1 - sin²θ

This statement is not true

D. sec²θ - 1 = 1/cot²θ

sec²θ - 1 = 1/cos²θ - 1

sec²θ - 1 = (1-cos²θ)/cos²θ

sec²θ - 1 = sin²θ/cos²θ

sec²θ - 1 = tan²θ

This statement is not true

E. sinθ cscθ = tan θ

sinθ cscθ = tan θ

sinθ cscθ = sinθ (1/sinθ)

sinθ cscθ = 1

This statement is not true

Therefore, the true statement is cosθ = cotθ/cscθ

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7 0

2 years ago

What is the solution to the system of equations below?

Lelechka [254]

Answer:

D

Step-by-step explanation:

equation form: x=4, y=5

3 0

4 years ago

What is the ratio of yellow sections to total sections, in the lowest terms?

Dovator [93]

It is choice a: 5/16

5 0

3 years ago

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The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6

IrinaK [193]

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$ , in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$ .

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$ .

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$ .

This is P(X = 4). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0

3 years ago

I NEED HELP!! WILL MARK BRAINLIEST!!!!

mixas84 [53]

Answer:

$-\frac{1}{32}$

the answer is A

Step-by-step explanation:

8 0

3 years ago

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