Write a recursive formula for each sequence. 11,5,-1,-7...

A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

Oksana_A [137]

Answer:

Step-by-step explanation:

Given that a bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7.

From the above we cannot conclude that red cards and even numbers are mutually exclusive

Just drawing two red cards and because the two happen to be odd we cannot generalize the red cards have odd numbers.

This might have occurred due to simple chance from a comparatively large number of 40 cards.

Suppose say we have red cards 20, and 19 red 1 blue.

Then drawing 2 from 19 red cards have more probability and this can occur by chance.

So friend's conclusion is wrong.

5 0

3 years ago

Find the distance between the pair of points given on the graph

seraphim [82]

The right answer is √74 units.

Please see the attached picture for full solution

Hope it helps

Good luck on your assignment

7 0

3 years ago

Solve for y 20x+5y=60 -20x -20x 5y=-20x+60 is it positive or negative 20?

valkas [14]

Step-by-step explanation:

it is negative 20 I'm pre sure

6 0

3 years ago

Please Help this is Due in 1 hour and I really dont understand any of this. (Question in Image)

Lerok [7]

Answer:

(-4, -1/2)

Step-by-step explanation:

Just by looking at the graph I can estimate that the midpoint of AB is (-4, -1/2) but we can also use midpoint formula to get a for sure answer.

Midpoint Formula: ((x1 + x2)/2, (y1 + y2)/2)

x: (-5 + (-3))/2

-8 / 2

x = -4

y: (-4 + 3)/2

-1 / 2

y = -1/2

6 0

3 years ago

Write the equation of the quadratic function whose graph passes through <img src="https://tex.z-dn.net/?f=%28-3%2C2%29" id="TexF

blagie [28]

Answer:

$f(x)=x^2+3x+2$

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

$f(x)=ax^2+bx+c$

Since it passes through the point (-3, 2). This means that when $x=-3$ , $f(x)=f(-3)=2$ . Hence:

$f(-3)=2=a(-3)^2+b(-3)+c$

Simplify:

$2=9a-3b+c$

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

$0=a(-1)^2+b(-1)+c \\ \\0=a-b+c$

And for (1, 6):

$6=a(1)^2+b(1)+c\\\\ 6=a+b+c$

So, we have a triple system of equations:

$\left\{ \begin{array}{ll} 2=9a-3b+c &\\ 0=a-b+c \\6=a+b+c \end{array} \right.$

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

$(0+6)=(a+a)+(-b+b)+(c+c)$

Compute:

$6=2a+2c$

Let's divide both sides by 2:

$3=a+c$

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

$-3(0)=-3(a-b+c)$

Multiply:

$0=-3a+3b-3c$

Add this to Equation 1:

$(0+2)=(9a-3a)+(-3b+3b)+(c-3c)$

Compute:

$2=6a-2c$

Again, we can divide both sides by 2:

$1=3a-c$

So, we know have two equations with only two variables:

$3=a+c\text{ and } 1=3a-c$

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

$(3+1)=(a+3a)+(c-c)$

Compute:

$4=4a$

Solve for a:

$a=1$

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

$3=a+c$

Substitute 1 for a and solve for c:

$\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}$

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

$6=a+b+c$

Substitute in known values and solve for b:

$6=(1)+b+(2)\\\\6=3+b\\\\b=3$

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

$f(x)=x^2+3x+2$

5 0

3 years ago