Ask question

Ask question

All categories

3 years ago

6

In a game between two teams a and b, john is set to win 200 dollars if a wins, and 100 dollars if b wins. john calculates his ex

pected winning to be 170 dollars. according to john's calculations, what is the probability that b wins (given no probability of a tied/drawn game)?

1 answer:

My name is Ann [436]3 years ago

7 0

According to his calculations, b has a 3/10 chance to win

You might be interested in

Can someone tell me whats1+1

Andreyy89

Answer:

2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A cylinder has a radius of 6 inches and a height of 15 inches. What is the surface area of the cylinder? Express the answer in t

shutvik [7]

Answer:

The surface area for this cylinder is 252*pi square inches.

Step-by-step explanation:

The surface area of a cylinder can be calculated by using the following formula:

surface area = 2*pi*r² + 2*pi*r*h

Applying the data from the problem, we have:

surface area = 2*pi*(6)² + 2*pi*(6)*15

surface area = 2*pi*36 + 180*pi

surface area = 72*pi + 180*pi

surface area = 252*pi

The surface area for this cylinder is 252*pi square inches.

4 0

3 years ago

Read 2 more answers

Is (4,4) a solution of the graphed system of inequalities? Choose 1 answer: Yes No

MissTica

No it is not. They are the wrong points

3 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

vodka [1.7K]

Answer:

B- In Step 2, the associative property cannot regroup addition and multiplication.

8 0

3 years ago

Read 2 more answers