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Calculate the mean for 31 29 33 37 43 38 33 40

Natasha2012 [34]

Answer:

The mode is 3 in the left and 3 in the right

Step-by-step explanation:

4 0

3 years ago

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Write 7/10 as a sum of fractions two different ways

Ilia_Sergeevich [38]

1/10+1/10+1/10+1/10+1/10+1/10+1/10 and 3/10+4/10

3 0

4 years ago

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Find the area of the region defined by the region defined by the inequality 2|x| + 3|y-1| ≤ 6

EleoNora [17]

If $x$ and $y-1$ have the same sign, then either

$x>0,y>1 \implies 2|x| + 3|y-1| = 2x + 3(y-1)=6 \implies 2x + 3y = 9$

or

$x$

If $x$ and $y-1$ have opposite sign, then

$x>0,y$

or

$x1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9$

This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set

$R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}$

The area of $R$ is given by the double integral

$\displaystyle \iint_R dx\,dy$

To compute the area, change the variables to

$\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}$

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}$

with determinant $\det(J) = -\frac1{12}$ . Then the integral transforms to

$\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du$

which is 1/12 the area of a square with side length 12. Hence the integral evaluates to

$\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}$ .

5 0

2 years ago

The unit for density is a derived unit. True or False

puteri [66]

Answer:

true

Step-by-step explanation:

8 0

3 years ago

For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X>7

Serga [27]

Answer:

1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be approaching to a value $\infty$ or it should not give us the $\frac{0}{0}$ (zero by zero) form when the input is given to the function.

The value of function will depend on the denominator.

Now, let us consider the given functions one by one:

1. 5y+2

Here denominator is 1. So, it can not attain a value $\infty$ or $\frac{0}{0}$ (zero by zero) form

So, for all real numbers, the function is defined.

$2.\ \dfrac{18}{y}$

At y = 0, the value

$At\ y =0, \dfrac{18}{y} \rightarrow \infty$

So, the given function is defined for all real numbers except y = 0

$3.\ \dfrac{1}{x+7}$

Let us consider denominator:

x + 7 can be zero at a value x = -7

$At\ x =-7, \dfrac{1}{x+7} \rightarrow \infty$

So, the given function is defined for all real numbers except x = -7

$4.\ \dfrac{2b}{10-b}$

Let us consider denominator:

10-b can be zero at a value b = 10

$At\ b =10, \dfrac{2b}{10-b} \rightarrow \infty$

So, the given function is defined for all real numbers except b = 10

7 0

3 years ago