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yaroslaw [1]
3 years ago
6

For this entry, you are a math expert at a local school. Explain to a group of students how independent and dependent events are

alike and how they are different.
Mathematics
1 answer:
GaryK [48]3 years ago
3 0

Answer:

hope this helps!

Step-by-step explanation:

Sample response: The two main variables in an experiment are the independent and dependent variable. An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.

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Identify each function as a constant, direct variation, absolute value, or greatest integer function.
Anton [14]
None of those describe this function.

It's not a constant function, because it is not equal to a constant (something that isn't a variable)

It doesn't show direct variation, because it can't be represented in the form
y = kx.

It isn't an absolute value function because the absolute value isn't taken anywhere.
If it was an absolute value function it would look something like this:
g(x) = |x+3|

It isn't a greatest integer function, because the greatest integer isn't taken anywhere.
If it was a greatest integer function it would look something like this:
g(x) = \left \lfloor{x + 3}\right \rfloor

Are you sure you typed that question right?
7 0
3 years ago
The midpoint of AB is M(1, 3). If the coordinates of A are(4,1), what are the coordinates of B?
Oxana [17]

Answer:

B(-2,5)

Step-by-step explanation:

\frac{x + 4}{2}  = 1 \\ 2 \times  \frac{x + 4}{2}  = 1 \times 2 \\ x + 4 = 2 \\ x = 2 - 4 \\ x =  - 2 \\  \frac{y + 1}{2}  = 3 \\ 2 \times  \frac{y + 1}{2}  = 3 \times 2 \\ y + 1 = 6 \\ y = 6  -  1 \\ y = 5

Hope that this is helpful.

Have a nice day.

6 0
2 years ago
Is 262015 divisible by 4?
eduard

Answer:

65503.75

Step-by-step explanation:

262015 divisble by 4 = 65503.75

4 0
3 years ago
Read 2 more answers
Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1
algol [13]

Find the critical points of f(x,y):

\dfrac{\partial f}{\partial x}=-2x=0\implies x=0

\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12

All three points lie within D, and f takes on values of

\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}

Now check for extrema on the boundary of D. Convert to polar coordinates:

f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t

Find the critical points of g(t):

\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0

\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2

\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi

where n is any integer. There are some redundant critical points, so we'll just consider 0\le t< 2\pi, which gives

t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3

which gives values of

\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}

So altogether, f(x,y) has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0
3 years ago
a language arts test is worth 100 points there is a total of 26 questions . there are spelling word questions that ae worth 2 po
natka813 [3]

10 question of spelling words and 16 questions of vocabulary are present

<em><u>Solution:</u></em>

Let "x" be the number of spelling word questions

Let "y" be the number of vocabulary word question

<em><u>There is a total of 26 questions. Therefore, we get</u></em>

number of spelling word questions + number of vocabulary word question = 26

x + y = 26 --------- eqn 1

<em><u>There are spelling word questions that worth 2 points each and vocabulary word question worth 5 points each</u></em>

The language arts test is worth 100 points

Therefore, we frame a equation as:

number of spelling word questions x 2 + number of vocabulary word question x 5 = 100

x \times 2 + y \times 5 = 100

2x + 5y = 100 --------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

x = 26 - y ------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

2(26 - y) + 5y = 100

52 - 2y + 5y = 100

3y = 100 - 52

3y = 48

<h3>y = 16</h3>

<em><u>Substitute y = 16 in eqn 3</u></em>

x = 26 - 16

<h3>x = 10</h3>

Thus 10 question of spelling words and 16 questions of vocabulary are present

4 0
3 years ago
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