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3 years ago

30 POINTS (Branliest) - Please Help

Mathematics

2 answers:

katrin2010 [14]3 years ago

8 0

Answer:

The answer is C

Step-by-step explanation:

I took the test

Ulleksa [173]3 years ago

3 0

Variability is the measure of the variation of a data set from the center. The mean and the median are the qualities that measure the center of a distribution.

For the first option, if more data is clustered around the median, (the measure of center), then the distribution has less variability than when less data is clustered around the median. Therefore, the first answer option is not correct because if more data is clustered around the median in the shaded plants set, then the shaded plant set will have less variability.

For the second option, the range is a measure of variability which measures the spread of the data set from the least value to the greatest value, but it does not take into account the variability of the other data values of the data set. Thus the range is regarded as a weak measure of variability and is not used when other measures of variability are available. Therefore, the second option is not correct.

For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the full sun plants set has a greater IQR, this shows that they have greater variability. Therefore, the third option is the correct answer.

For the fourth option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the fourth option is not the correct answer.

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Which expression equals -(9x + 8) - 2(x -5)

Troyanec [42]

-(9x + 8) - 2(x -5)
= -9x - 8 - 2x + 10
= -11x + 2

3 0

3 years ago

Read 2 more answers

Find the slope of the line that passes through (-10, -52) and (56, -37).

hichkok12 [17]

15/66 I think you might be able to simplify

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2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

Third-degree, with zeros of −3 , − 2 , and 1 , and passes through the point ( 3 , 11 ) .

Masja [62]

Answer:

y = ( x + 3 ) * ( x + 2 ) * ( x - 1 )* 11/60

Step-by-step explanation:

Lets y = f(x) in a cartesian coordinates

Having three zeros:

x = -3 ⇒ x = -2 ⇒ and x = 1

That meas that if x takes the above mentioned values " y " must be zero

therefore y must be of the form

f (x) = y = ( x + 3 ) * ( x + 2 ) * ( x - 1 ) (1)

In that case for y to be zero one of the factors should be zero or

y = 0 x + 3 = 0 and x = - 3 is a zero of the function y .

The same reasoning applies for the other two roots

Now we have to evaluate the other condition.

According to problem statement the function passes through the point ( 3, 11 ) , that means that when x = 3 , y have to be 11, therefore we plug in equation (1) that value to see what happens

y = ( x + 3 ) * ( x + 2 ) * ( x - 1 )

11 = ( 3 + 3 ) * ( 3 + 2 ) + ( 3 - 1) = 6*5*2 = 60

Then we adjust the expression (1) to meet the condition of the function passing through point ( 3 , 11) as:

y = ( 3 + 3 ) * ( 3 + 2 ) + ( 3 - 1) * 11/60 (2)

and check to see if we did right

for y to be zero x can be x = - 3 x = -2 and x = 1 in all these cases y = 0. And if x = 3 in equation (2) y = 11. And that what we want to shown. Then the solution is:

y = ( x + 3 ) * ( x + 2 ) * ( x - 1 )* 11/60

8 0

4 years ago

297 is the sum of F and 211 write equation (IXL)

Lelechka [254]

297 is the sum of F and 211

297 = f+211

7 0

3 years ago