Ask question

Ask question

All categories

2 years ago

8

Select all that apply. If there are 8 chocolate chip cookies out of 20 total cookies in a jar, what is the probability that you

will randomly choose a chocolate chip cookie?
A. 2/5
B. 0.4
C. 0.8
D. 40%

2 answers:

Gelneren [198K]2 years ago

7 0

Answer:

B) 0.4, D) 40% and A) 2/5

Step-by-step explanation:

Part/Whole=%/100

8/20=0.4*100=40%=2/5=0.4

marysya [2.9K]2 years ago

5 0

Answer:

A, B, D

Step-by-step explanation:

8 divided by 20

You might be interested in

Simplify the following expression (62)4

bija089 [108]

Answer:

$3^{0} =1$

Step-by-step explanation:

7 0

3 years ago

Enter the values to complete the input/output table for the function.

Artist 52 [7]

Given function : 3x−6y=12.

We are given x : −2 0 4.

We need to find the values of y's for x=-2, x=0 and x=4.

Plugging x=-2 in the given equation, we get

3(-2)−6y=12

-6 - 6y = 12.

Adding 6 on both sides, we get

-6+6 - 6y = 12+6

-6y = 18.

Dividing by -6 on both sides, we get

y= -3.

On the same way, plugging x=0.

3(0)−6y=12

-6y =12.

y=-2.

Plugging x=4,

3(4)−6y=12

12 -6y = 12.

Subtracting 12 on both sides.

12-12 -6y = 12-12

-6y=0

y=0.

Therefore,

<h3>x −2 0 4</h3><h3>y -3 -2 0</h3>

3 0

3 years ago

If the graph of Rx) = |x| is shifted up 7 units, which equation represents the new graph? OA. g(x) = (x + 7|| O B. g(x) = |x-71

Tju [1.3M]

Answer:

answer B

Step-by-step explanation:

hope this helps

8 0

3 years ago

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago

Just need to know what x is haha

marishachu [46]

Answer: x=14

Step-by-step explanation:

The angles in a triangle are equal to 180 so

6x + 3 + 5x + x + 9 = 180

12x + 12 = 180

-12 -12

12x = 168

x = 14

4 0

3 years ago