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elixir [45]
3 years ago
13

2. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R), or goes

straight ahead (A). The experiment terminates as soon as a car is observed to turn left. Let X= the number of cars observed. What are possible X values?
Given the outcomes below, find their associated X values.
Outcome: RRL AARRL AARL RRAL ARL
x= ____ ____ ____ ____ ____
Mathematics
1 answer:
jasenka [17]3 years ago
7 0

Answer:

Natural numbers (integers greater than zero)

X = 3,  5,  4,  4,  3

Step-by-step explanation:

The least number of cars that can be observed in this experiment is 1, if the first car turns left. On the other hand, the experiment could go on forever if no car ever turns left, thus the highest number of cars approaches infinite.

The possible values of X are integers greater than zero, which are known as the Natural numbers.

If X = number of cars observed, simply count the number of letters in each outcome for the value of X:

Outcome = RRL, AARRL, AARL, RRAL, ARL

            X = 3,  5,  4,  4,  3

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Step-by-step explanation:

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3 years ago
A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2
andrey2020 [161]

Answer:

The required position of the particle at time t is: x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}

Step-by-step explanation:

Consider the provided matrix.

v_1=\begin{bmatrix}-3\\1 \end{bmatrix}

v_2=\begin{bmatrix}-1\\1 \end{bmatrix}

\lambda_1=4, \lambda_2=2

The general solution of the equation x'=Ax

x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}

Substitute the respective values we get:

x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}

x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}

Substitute initial condition x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}

\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}

Reduce matrix to reduced row echelon form.

\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}

Therefore, c_1=2.5,c_2=1.5

Thus, the general solution of the equation x'=Ax

x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}

x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}

The required position of the particle at time t is: x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}

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3 years ago
A veterinarian will prescribe an antibiotic to a dog based on its weight. The effective dosage of the antibiotic is given by d ≥
gregori [183]

Answer:

B) (35, 260)

Step-by-step explanation:

A veterinarian will prescribe an antibiotic to a dog based on its weight. The effective dosage of the antibiotic is given by d ≥ 1∕5w2, where d is dosage in milligrams and w is the dog's weight in pounds. Which of the following ordered pairs gives an effective dosage of antibiotics for a 35-pound dog?

A) (35, 240)

B) (35, 260)

C) (260, 35)

D) (240, 35)

Ordered pairs is composed of pairs, usually an x coordinate and a y coordinate. It refers to a location of a point on the coordinate. It matches numbers to functions or relations.

Given the relation between d is dosage in milligrams and w is the dog's weight in pounds as d ≥ 1∕5w²

For a 35 pound dog (i.e w = 35 pound). The dosage is given as:

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For an ordered pair (x, y), x is the independent variable (input) and y is the dependent variable (output).

The dog weight is the independent variable and the dosage is the dependent variable.

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Gelneren [198K]

The value of d from the given expression is 15

<h3>Word problem leading to equation</h3>

From the given statement the sum of five times d and five times five is a

hundred is interpreted as;

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Subtract 25 from both sides

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Learn more on word problem here: brainly.com/question/13818690

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See attached picture:

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