Ask question

Ask question

All categories

2 years ago

7

You're ordering a sandwich at the local deli, below is a tree diagram of the possible orders. What is the probability that your

sandwich has turkey AND swiss cheese?

1 answer:

dusya [7]2 years ago

4 0

Answer:

Step-by-step explanation:

1. So there is two bread options but let's ignore that due every option being the same for both bread options.

2. So if you look there are three meat options meaning there is a one in three chance logical chance someone picks turkey (note when I say logical I mean that doesn't mean that this will be true in real life) and there is a one in two logical chance someone picks Swiss cheese.

3. So with a one in three chance logical meat option and a one two chance cheese option and.

4. Now knowing this it is simple math at this point as simple as $\frac{1}{3}$ x $\frac{1}{2}$

5. 1 x 1 = 1 and 3 x 2 = 6

6. leading to $\frac{1}{6}$

You might be interested in

HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider<im

GrogVix [38]

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for $f:\mathbb R^2\to\mathbb R$ , we have the partial derivative with respect to $x$ defined as

$\displaystyle \frac{\partial f}{\partial x} = \lim_{h\to0}\frac{f(x+h,y) - f(x,y)}h$

The derivative at (0, 0) is then

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(0+h,0) - f(0,0)}h$

• By definition of $f$ , $f(0,0)=0$ , so

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)}h = \lim_{h\to0}\frac{\tan^2(g(h,0))}{h\cdot g(h,0)}$

• Expanding the tangent in terms of sine and cosine gives

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{h\cdot g(h,0) \cdot \cos^2(g(h,0))}$

• Introduce a factor of $g(h,0)$ in the numerator, then distribute the limit over the resulting product as

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{\sin^2(g(h,0))}{g(h,0)^2} \cdot \lim_{h\to0}\frac1{\cos^2(g(h,0))} \cdot \lim_{h\to0}\frac{g(h,0)}h$

• The first limit is 1; recall that for $a\neq0$ , we have

$\displaystyle\lim_{x\to0}\frac{\sin(ax)}{ax}=1$

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)}h = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h$

and this is exactly the partial derivative of $g$ with respect to $x$ .

$\displaystyle \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{g(h,0)-g(0,0)}h = \frac{\partial g}{\partial x}(0,0)$

For the same reasons shown above,

$\displaystyle \frac{\partial f}{\partial y}(0,0) = \frac{\partial g}{\partial y}(0,0)$

(b) To show that $f$ is differentiable at (0, 0), we first need to show that $f$ is continuous.

• By definition of continuity, we need to show that

$\left|f(x,y)-f(0,0)\right|$

is very small, and that as we move the point $(x,y)$ closer to the origin, $f(x,y)$ converges to $f(0,0)$ .

We have

$\left|f(x,y)-f(0,0)\right| = \left|\dfrac{\tan^2(g(x,y))}{g(x,y)}\right| \\\\ = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)^2}\cdot\dfrac{g(x,y)}{\cos^2(g(x,y))}\right| \\\\ = \left|\dfrac{\sin(g(x,y))}{g(x,y)}\right|^2 \cdot \dfrac{|g(x,y)|}{\cos^2(x,y)}$

The first expression in the product is bounded above by 1, since $|\sin(x)|\le|x|$ for all $x$ . Then as $(x,y)$ approaches the origin,

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{|g(x,y)|}{\cos^2(x,y)} = 0$

So, $f$ is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which $f(x,y)$ changes as we move the point $(x,y)$ closer to the origin, given by

$\left|\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\right|,$

approaches 0.

Just like before,

$\left|\dfrac{\tan^2(g(x,y))}{g(x,y)\sqrt{x^2+y^2}}\right| = \left|\dfrac{\sin^2(g(x,y))}{g(x,y)}\right|^2 \cdot \left|\dfrac{g(x,y)}{\cos^2(g(x,y))\sqrt{x^2+y^2}}\right| \\\\ \le \dfrac{|g(x,y)|}{\cos^2(g(x,y))\sqrt{x^2+y^2}}$

and this converges to $g(0,0)=0$ , since differentiability of $g$ means

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{g(x,y)-g(0,0)}{\sqrt{x^2+y^2}}=0$

So, $f$ is differentiable at (0, 0).

3 0

3 years ago

Write is the formula for the discriminant?

Svetlanka [38]

Answer:

The number D = b2 – 4ac determined from the coefficients of the equation ax2 + bx + c = 0. The discriminant reveals what type of roots the equation has. Note: b2 – 4ac comes from the quadratic formula.

Hope this helps please mark me as brainliest

5 0

3 years ago

Read 2 more answers

1. A partially completed chart shows the hierarchy of a set of polygons.

lana [24]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Please find the attached file for the complete solution.

5 0

3 years ago

Which factorizations can be used to identify the real zeros of the function f(x)=20x^2+23x-6

Ainat [17]

It’s the quadratic formula because it’s give directly the zeros of a function

5 0

4 years ago

Greatest to least 1/4, 0.08, 3/7, 82%

JulijaS [17]

Answer:

.08, .25, .42, .82

Step-by-step explanation:

1/4, 0.08, 3/7, 82%

.25, 0.08, .42, .82

6 0

3 years ago