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2 years ago

Use linear regression to find a function that fits the following points. (-5,14), (1,-16)

Mathematics

1 answer:

Scrat [10]2 years ago

7 0

The linear regression equation is y = -5x + 21

<h3>How to determine the equation?</h3>

The points are given as:

(-5,14), (1,-16)

The equation is then calculated using:

$y = \frac{y_2 -y_1}{x_2 -x_1} * (x - x_1) + y_1$

This gives

$y = \frac{14 + 16}{-5 - 1} * (x - 1) + 16$

Evaluate

y = -5 (x - 1) + 16

Expand

y = -5x + 5 + 16

Evaluate the sum

y = -5x + 21

Hence, the linear regression equation is y = -5x + 21

Read more about linear regression at:

brainly.com/question/17844286

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3 years ago

What is the measure of the angle

Paul [167]

Answer:

Option A, 50 degrees

Step-by-step explanation:

<u>Step 1: Estimate this angle</u>

<em>Option A, works because looks about the size of 50 degrees</em>

Option B, doesn't work because it is not more than a right angle.

Option C, doesn't work because it is not an obtuse angle.

Option D, doesn't work because there is no way it is 15.

Answer: Option A, 50 degrees

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4 years ago

"What is the probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice? Assume you’re throwing a single d

dem82 [27]

Answer:

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

Step-by-step explanation:

For each throw, there are only two possible outcomes. Either it is a '3', or it is not. This means that we solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

There are 6 possible outcomes for the dice. This means that the probability that it is a '3' is $\frac{1}{6} = 0.167$

There are 10 throws, so $n = 10$ .

Probability of throwing AT LEAST two ‘3’s on the dice?

Either you throw less than two, or you throw at least two. The sum of the probabilities of these events is 1. So

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

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

In which

$P(X < 2) = P(X = 0) + P(X = 1)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.167)^{0}.(0.833)^{10} = 0.1609$

$P(X = 1) = C_{10,1}.(0.167)^{1}.(0.833)^{9} = 0.3225$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1609 + 0.3225 = 0.4834$ .

Finally:

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4839 = 0.5161$ .

There is a 51.61% probability that in 10 dice throws, you will throw AT LEAST two ‘3’s on the dice.

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3 years ago

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