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4 years ago

1. Identifique a abscissa e a ordenada dos pontos abaixo.

Mathematics

2 answers:

Arte-miy333 [17]4 years ago

4 0

Answer:

A(3,-5) abscissa:3 ordenada: -5

B(-1,0) abscissa: -1 ordenada: 0

C(-3,5;-2) abscissa: -3.5 ordenada -2

D(0,-1) abscisa: 0 ordenada:- 1

Step-by-step explanation:

What we must take into account is that the abscissa is the value of x and the ordinate is the value of y. There is always a number of the (x,y), that is, the abscissa is the first value and the ordinate is the second value, therefore:

trasher [3.6K]4 years ago

4 0

Answer:

We need to identify abscissa and ordenada in each of the pairs given.

For a pair (x, y), x is the abscissa and y is the ordenada.

A (3,-5) abscissa_____3____ ordenada______-5_______

B (-1,0) abscissa____-1____ ordenada_______0_____

C (-3,5;-2) abscissa_____-3.5____ ordenada_____-2______

D (0,-1) abscissa_____0___ ordenada___-1_______

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Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

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