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Studentka2010 [4]

4 years ago

5

About 13 out of 20 homes have a personal computer. On a street with 60 homes, how many would you except to have a personal compu

ter?

1 answer:

melamori03 [73]4 years ago

7 0

Out of the 20 homes, only 13 have a personal computer.

20 goes into 60 how many times?
60/20=3

Multiply 13 by 3 because that is how many "groups" there are
13*3=39

There are 39 houses with computers

You might be interested in

Nina knows that the average of the x-intercepts represents the line of symmetry for a quadratic function through the x-axis. Whi

Dafna1 [17]

To get the x-intercepts of the function, f(x) = 4x2 – 24x + 20

It has to be equated to zero and the values of x are the x-intercepts. So,
4x2 – 24x + 20 = 0
The resulting equation is a quadratic equation which can be solved by different methods. The solution is
x = 5, 1
The average therefore is:
(1+5/)2 = 3

6 0

3 years ago

Read 2 more answers

Number patterns 54, 18, 6, 2, 2/3, what are the next 2 terms?

nikitadnepr [17]

Answer:

2/9, 2/27

Step-by-step explanation:

54, 18, 6, 2, 2/3

Take the second term and divide by the first

18/54 = 1/3

We multiply by 1/3

Check by taking the third term and dividing by the second

6/18 = 1/3

This works

Take the last term and multiply by 1/3

2/3*1/3 = 2/9

2/9*1/3 =2/27

4 0

3 years ago

How is 5/2 divided by 1/4 equals 10 responsible

vredina [299]

Answer:

Step-by-step explanation:

'll just help with proving whether she is correct.

First of all, a fraction divided by a fraction can be made into multiplication since the division of two fractions is kinda difficult

so remember keep, change, flip

keep the first fractions, change the division symbol into multiplication, then flip the fraction i.e. 1/2 becomes 2/1

so 5/2 divided by 1/4 becomes

5/2 * 4/1

which then solved becomes

20/2 which is simplified to 10

Hope this helps :)

8 0

4 years ago

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose verti

mojhsa [17]

$D$ is the set of points,

$\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}$

Check for critical points:

$f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}$

Of these 4 points, only 2 belong to $D$ , (-1, 2) and (1, 2), for which we have

$\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}$

Now look for extrema along the boundary.

If $x=-2$ , then

$f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2$

We have $f'(-2,y)>0$ for $-2$ and $f'(-2,y)$ for $2$ , which indicates a local maximum at $y=2$ and minima at the endpoints of this boundary. So

$\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}$

If $x=2$ , then

$f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2$

We have $f'(2,y)$ for $2$ , so we have extrema at the endpoints of this boundary.

$\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}$

If $y=x$ , then

$f(x,x)=9x+1\implies f'(x,x)=9>0$

which tells us $f$ is strictly increasing on this boundary, giving the extrema we already know about,

$\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}$

If $y=3$ , then

$f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1$

We have $f'(x,3)>0$ for $-2$ and $1$ , and $f'(x,3)$ for $-1$ . This indicates a maximum at $x=-1$ and a minimum at $x=1$ , with

$\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}$

From this analysis, we find that $f$ attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).

4 0

3 years ago

How do you find the difference between median and mean

Flura [38]

Median is the middle number. Ex:
1,3,5,7,10,13,14

The middle number is 7 because there is three numbers on either side of 7. The median would be 7.

The mean is when you add up all of the numbers and divide by the number of numbers. For example, 1,3,5,7,10,13,14
You add all of them. 1+3+5+7+10+13+14 equals 53. Then divide 53 by 7 because there are seven digits.

Median equals 7.57142857

5 0

4 years ago