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galben [10]
3 years ago
15

(0, 3) and (2,5) point slope form

Mathematics
2 answers:
snow_lady [41]3 years ago
8 0
<h3>PROBLEM</h3>

0-5/ 3-2

0-5=-5  

3-2= 1

-5/1 or -5

ANSWER

-5

Anastasy [175]3 years ago
6 0

Answer:

The slope intercept form is..

y = x + 3

The slope is m = 1

Did any of these help you answer the question?

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3x+3

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Elodia [21]

All exercises involve the same concept, so I'll show you how to do the first, then you can apply the exact same logic to all the others.

The first thing you need to know is that, when a certain quantity multiplies a parenthesis, you can distribute that number to every element in the parenthesis. This means that

a(b+c) = ab+ac

So, a is multiplying the parenthesis involving b and c, and we distributed it: a multiplies both  b and c in the final result.

Secondly, you have to know how to recognize like terms, because they are the only terms you can sum. Two terms can be summed if they have the same literal expression. So, for example, you cannot sum 3x+2y, and neither 5x^2-4x exponents count.

But you can su, for example,

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or

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So, take for example exercise 9:

1.2(7x-5y) - (10y-4.3x)

We distribute the 1.2 through the first parenthesis:

1.2(7x-5y) = 1.2 \cdot 7x - 1.2 \cdot 5y = 8.4x-6y

And you can distribute the negative sign through the second parenthesis (it counts as a -1 to distribute):

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A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li
svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

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}.\pi^{x}.(1-\pi)^{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 \pi is the probability of X happening.

In this problem

15 drivers are randomly selected, so n = 15.

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year.  This means that \pi = 0.08.

What is the probability of getting 3 or more who were involved in a car accident last year?

This is P(X \geq 3).

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

P(X < 3) + P(X \geq 3) = 1

P(X \geq 3) = 1 - P(X < 3)

In which

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{15,0}.(0.08)^{0}.(0.92)^{15} = 0.2863

P(X = 1) = C_{15,1}.(0.08)^{1}.(0.92)^{14} = 0.3734

P(X = 2) = C_{15,2}.(0.08)^{2}.(0.92)^{13} = 0.2273

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2863 + 0.3734 + 0.2273 = 0.887

Finally

P(X \geq 3) = 1 - P(X < 3) = 1 - 0.887 = 0.113

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

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