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3 years ago

Compare each set of data. Determine which set shows greater variability.

Mathematics

2 answers:

Nadya [2.5K]3 years ago

9 0

Answer:

Which set shows greater variability? Set B

Which measure provides greater variability for the set you chose? Range

alexdok [17]3 years ago

3 0

Answer:hhfjb

Step-by-step explanation:hdbbcjb

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Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

Point a:

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

Point b:

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for y we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$

Point c:

To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:

Find the values y(0.2), y(0.6), and y(1) using $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

$y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771$

$y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965$

$y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504$

Next, where $y_{1}, y_{3}, \:and \:y_{5}$ are from the table.

$|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771$

$|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796$

$|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859$

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3 years ago

What is the same as 1 cm to 1 km

lana [24]

Pretty easy little one hehe, hoped this helped 1m=1000mm

6 0

2 years ago

Find a numerical value if one trigonometric function of x if tan2x-sin2x/sin2x=5

seropon [69]

Answer:

The values of x that satisfy the given equation are:

x1 = 1.183 + nπ

x2 = -1.183 + nπ

Step-by-step explanation:

Given tan²x - sin²x/sin²x = 5

Simplifying this, we have

tan²x - 1 = 5

Adding 1 to both sides, we have

tan²x = 6

Because tan²x = (tanx)², we can write as

(tanx)² = 6

Taking square roots of both sides, we have.

tanx = ±√6

x = arctan(±√6) + nπ

≈ 1.183 + nπ or -1.183 + nπ

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2 years ago

How to find the median on a dot plot?

pickupchik [31]

Answer:

In order to find the median of the data you take every value and order it from least to greatest. When you do this make sure you write a number the correct amount of times. for example if there are 4 dots above 6 on the dot plot then you need to make sure you write 6 4 times. After you have all the data from least to greatest find the number in the middle. that is the median.

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3 years ago

The hypotenuse of a 45°-45°-90° triangle measures 22 2 units.

Aleonysh [2.5K]

Answer:

Option C.

Step-by-step explanation:

Given information: The hypotenuse of a 45°-45°-90° triangle measures 22√2 units.

Let x be the length of one leg.

From the given figure it is clear that length of both legs are same.

According to the Pythagoras theorem, in a right angled triangle

$(leg_1)^2+(leg_2)^2=hypotenuse^2$