The following set of random numbers represents 20 simulations of three daily

Mathematics

1 answer:

babunello [35]2 years ago

3 0

The number of simulations in which was there was an on-time departure for all three flights is; 6

<h3>How to carry out number simulations?</h3>

We are given;

Number of Simulations = 20

Numbers that represent late departure = 0, 1, 2 or 3

Numbers that represent On-time departure = 4, 5, 6, 7, 8 or 9.

Now, the 20 simulations as seen online are;

339 931 317 444 432 858 194 192 649 709

742 626 702 258 955 880 505 854 774 132

To know the number of simulations that were on-time departure, we need to find the ones that don't include 0, 1, 2 or 3.

Looking at the simulations, the ones that represent on-time departure are 444, 858, 649, 955, 854, 774.

Thus, there are 6 simulations in which was there an on-time departure for all three flights

Read more about Number Simulations at; brainly.com/question/1459420

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What is an equation of the line that passes through the points (0, 3) and (5,-3)?

Arisa [49]

Answer:

The equation of the line that passes through the points (0, 3) and (5, -3) is $y = -\frac{6}{5}\cdot x +3$ .

Step-by-step explanation:

From Analytical Geometry we must remember that a line can be formed after knowing two distinct points on Cartesian plane. The equation of the line is described below:

$y = m\cdot x + b$ (Eq. 1)

Where:

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

$m$ - Slope, dimensionless.

$b$ - y-Intercept, dimensionless.

If we know that $(x_{1},y_{1}) = (0,3)$ and $(x_{2},y_{2})=(5,-3)$ , the following system of linear equations is constructed:

$b = 3$ (Eq. 2)

$5\cdot m + b = -3$ (Eq. 3)

The solution of the system is: $b = 3$ , $m = -\frac{6}{5}$ . Hence, we get that equation of the line that passes through the points (0, 3) and (5, -3) is $y = -\frac{6}{5}\cdot x +3$ .