Asap help please on this question thanks

Which of the following best describes the relationship between the data displayed in the scatter plot below?

Naya [18.7K]

Answer: No association

6 0

2 years ago

What is 0.433 rounded to the nearest tenths

Fynjy0 [20]

0.433 is already rounded to the hundredths so to round it to the nearest tenths, remove the last number.

Your answer is 0.43

6 0

2 years ago

Read 2 more answers

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

2 years ago

Are these answers correct?

Nikolay [14]

Answer:

You're correct.

Step-by-step explanation:

3 0

2 years ago

The linear function f(x) = 0.9x + 79 represents the average test score in your math class, where x is the number of the test tak

Levart [38]

Given:

The linear function for average test score in your math class is

$f(x)=0.9x+79$

where x is the number of the test taken.

The linear function g(x) in the given table represents the average test score in your science class, where x is the number of the test taken.

To find:

Part A: The test average for your math class after completing test 2.

Part B: The test average for your science class after completing test 2.

Part C: Which class had a higher average after completing test 4?

Solution:

Part A:

We have,

$f(x)=0.9x+79$

Put x=2 in the above function, to find the test average for your math class after completing test 2.

$f(2)=0.9(2)+79$

$f(2)=1.8+79$

$f(2)=80.8$

Therefore, the test average for your math class after completing test 2 is 80.8.

Part B:

From the given table it is clear that g(x) =79 at x=2.

Therefore, the test average for your science class after completing test 2 is 79.

Part C:

Put x=4 in f(x), to find the test average for your math class after completing test 4.

$f(4)=0.9(4)+79$

$f(4)=3.6+79$

$f(4)=82.6$

From the table it is clear that the value of g(x) is increased by 1 as x is increased by 1. So, the value of function g(x) at x=4 must be 1 more than 80.

$g(4)=81$

It is clear that,

$82.6>81$

Therefore, math class had a higher average after completing test 4.

3 0

3 years ago