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3 years ago

Find BC. Round to the nearest tenth

Mathematics

2 answers:

Crazy boy [7]3 years ago

7 0

It’s 20.8 for anybody that’s confused

melamori03 [73]3 years ago

6 0

Use the law of sines; see the attached picture.

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Which of the following pairs of functions are inverses of each other?

GenaCL600 [577]

Answer:

I think its B

Step-by-step explanation:

6 0

2 years ago

NO LINKS!!! What is the transformation f(x)= x^3:

Mama L [17]

Answer:

4. Horizontal shrink by a factor of ¹/₅

5. Left 5, Up 5

6. Right 5, Down 5

Step-by-step explanation:

Transformations of Graphs (functions) is the process by which a function is moved or resized to produce a variation of the original (parent) function.

Transformations

For a > 0

$f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}$

$f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}$

$f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}$

$f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}$

$y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a$

$y=f(ax) \implies f(x) \: \textsf{stretched parallel to the x-axis (horizontally) by a factor of} \: \dfrac{1}{a}$

$y=-f(x) \implies f(x) \: \textsf{reflected in the} \: x \textsf{-axis}$

$y=f(-x) \implies f(x) \: \textsf{reflected in the} \: y \textsf{-axis}$

Identify the transformations that take the parent function to the given function.

Question 4

$\textsf{Parent function}: \quad f(x)=x^3$

$\textsf{Given function}: \quad f(x)=(5x)^3$

Comparing the parent function with the given function, we can see that the x-value of the parent function has been multiplied by 5.

Therefore, the transformation is:

$y=f(5x) \implies f(x) \: \textsf{stretched parallel to the x-axis (horizontally) by a factor of} \: \dfrac{1}{5}$

As a > 1, the transformation visually is a compression in the x-direction, so we can also say: Horizontal shrink by a factor of ¹/₅

Question 5

$\textsf{Parent function}: \quad f(x)=x^3$

$\textsf{Given function}: \quad f(x)=(x+5)^3+5$

Comparing the parent function with the given function, we can see that there are a series of transformations:

Step 1

5 has been added to the x-value of the parent function.

$f(x+5) \implies f(x) \: \textsf{translated}\:5\:\textsf{units left}$

Step 2

5 has then been added to function.

$f(x+5)+5 \implies f(x+5) \: \textsf{translated}\:5\:\textsf{units up}$

Transformation: Left 5, Up 5

Question 6

$\textsf{Parent function}: \quad f(x)=x^3$

$\textsf{Given function}: \quad f(x)=(x-5)^3-5$

Comparing the parent function with the given function, we can see that there are a series of transformations:

Step 1

5 has been subtracted from the x-value of the parent function.

$f(x-5) \implies f(x) \: \textsf{translated}\:5\:\textsf{units right}$

Step 2

5 has then been subtracted from function.

$f(x-5)-5 \implies f(x-5) \: \textsf{translated}\:5\:\textsf{units down}$

Transformation: Right 5, Down 5

Learn more about graph transformations here:

brainly.com/question/27845947

6 0

1 year ago

Read 2 more answers

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$