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3 years ago

What is an equation of the line that passes through the point (8,-1) and (4,-1)?

Mathematics

1 answer:

GalinKa [24]3 years ago

7 0

Answer:

Step-by-step explanation:

We can start with the general form of the equation for a straight line (may we assume this is a straight line?): y=mx+b. m is the slope (the rate of change of y for a change in x) and b is the y-intercept (the value of y when x=0).

Slope is the Rise of the line (the y axis) and Run is the change in the x axis to create the change in y. The slope is the Rise/Run, which we can find if we have two points on the line.

(8,-1) and (4,-1) will gives us the information to find the slope, m. The Rise is the change in y:

(-1-(-1)) = 0 !!!! It doesn't matter what the change in x is - the slope will be zero. It it weren't zero, it would be calculated as (8 - 4) or 4. The Rise/Run is 0/4 or 0 for this line. It is a flat line! The equation:

y=mx+b

y = 0x + b

y = b

This tells us the value of y will be the y-intercept, which is -1. Y is the same regardless of the value of x.

The equation is y=-4

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Which of the following functions has a rate of change that stays the same?

aalyn [17]

Answer:

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3 years ago

A cable car starts off with n riders. The times between successive stops of the car are independent exponential random variables

nikitadnepr [17]

Answer:

The distribution is $\frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}$

Solution:

As per the question:

Total no. of riders = n

Now, suppose the $T_{i}$ is the time between the departure of the rider i - 1 and i from the cable car.

where

$T_{i}$ = independent exponential random variable whose rate is $\lambda$

The general form is given by:

$T_{i} = \lambda e^{- lambda}$

(a) Now, the time distribution of the last rider is given as the sum total of the time of each rider:

$S_{n} = T_{1} + T_{2} + ........ + T_{n}$

$S_{n} = \sum_{i}^{n} T_{n}$

Now, the sum of the exponential random variable with $\lambda$ with rate $\lambda$ is given by:

$S_{n} = f(t:n, \lamda) = \frac{\lambda^{n}e^{- \lambda t}t^{n - 1}}{(n - 1)!}$

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3 years ago

Help please logarithms

DochEvi [55]

Answer:

Simplified: 40

Step-by-step explanation:

To simplify this, we need to first separate the terms in this expression. There is a property of exponents that says: $x^{a+b} =x^ax^b$ We can do this same thing to simplify this expression. Next, we can simplify each of these terms. $2^2$ simply becomes 4. In the other term, the $2^{log_2}$ simply cancel each other and leave the 10. This means that we're left with 4*10, which is 40