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coldgirl [10]
3 years ago
13

Fire disturbance of terrestrial biomes is one of the primary factors in non-native plant species growth. A 2014 study by Pec and

Carlton investigated the aftereffects of controlled fires on non-native grass growth. The data provided give measurements of the biomass in milligrams of non-native grass and native chaparral shrubs for several randomly selected plots of land in Bell Canyon, California. Click to download the data in your preferred format. CrunchIt! CSV Excel JMP Mac-Text Minitab PC-Text R SPSS TI-Calc Calculate the y-intercept and slope of the least-squares regression line using non-native grass biomass to predict chaparral shrubs biomass. Give your answers precise to three decimal places y-intercept mg slope0.401 Using the least-squares regression line, calculate the predicted value and residual value for the data point where the non- native grass biomass is equal to 57 and the chaparral shrubs biomass is equal to 27. Give your answers precise to three decimal places. predicted value: mg residual value: Select the statement that represents the correct interpretation of the residual value. OThe residual value measures the accuracy of the least-squares model. The residual value measures the variation in the response variable. The residual value measures how far away the observed value is from the predicted valueon_native_grass chaparral10 3726 399 6641 3763 2835 2757 2716 4123 5444 837 2650 2633 1343 1052 1541 2334 1597 567 693 072 1261 075 175 784 490 065 680 385 382 570 372 166 374 892 1473 1067 633 61 052 625 053 650 666 265 060 051 047 2250 857 21

Mathematics
2 answers:
Amiraneli [1.4K]3 years ago
7 0

Answer:

Step-by-step explanation:

answer is attached below

kolezko [41]3 years ago
7 0

Answer:

Step-by-step explanation:

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A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at th
nasty-shy [4]

Answer:

(9.6, 25.7) is a 80% confidence interval for the average net change in a student's score after completing the course.

Step-by-step explanation:

We have n = 6, \bar{x} =  17.6667 and s = 13.3367. The confidence interval is given by

\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}}) where t_{\alpha/2} is the \alpha/2th quantile of the t distribution with n-1=5 degrees of freedom. As we want the 80% confidence interval, we have that \alpha = 0.2 and the confidence interval is 17.6667\pm t_{0.1}(\frac{13.3367}{\sqrt{6}}) where t_{0.1} is the 10th quantile of the t distribution with 5 df, i.e., t_{0.1} = -1.4759. Then, we have 17.6667\pm (1.4759)(\frac{13.3367}{\sqrt{6}}) and the 80% confidence interval is given by (9.6, 25.7)

6 0
3 years ago
What is the missing number in this sequence:<br> 1, 16, __, 100, 169
Viefleur [7K]

Answer: 49

Step-by-step explanation:

7 0
4 years ago
What is the x-intercept of the graph 0f 27= -3x2?
jok3333 [9.3K]

Answer:

x=±√−9

No real solutions.

Step-by-step explanation:

Step 1: Add 3x^2 to both sides.

27+3x2=−3x2+3x2

3x2+27=0

Step 2: Subtract 27 from both sides.

3x2+27−27=0−27

3x2=−27

Step 3: Divide both sides by 3.

3x2 /3 = −27 /3

x2=−9

Step 4: Take square root.

x=±√−9

5 0
3 years ago
Compute the matrix of partial derivatives of the following functions.
s344n2d4d5 [400]

For a vector-valued function

\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))

the matrix of partial derivatives (a.k.a. the Jacobian) is the m\times n matrix in which the (i,j)-th entry is the derivative of f_i with respect to x_j:

D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}

So we have

(a)

D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}

(b)

D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}

(c)

Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}

(d)

Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}

5 0
3 years ago
Solve for x.<br> 2x + 20 2x - 4<br> #<br> X =<br> = [?]
Dima020 [189]

alr! so we know that the top line is straight, so it=180 degrees. then we know...

2x + 20 + 2x - 4= 180

4x + 16=180

4x=164

x=82

youre welcome!

4 0
3 years ago
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