Events d and e are independent, with p( d ) = 0. 6 and p( d and e ) = 0. 18. Which of the following is true?.

When you lose your bsf and you dont know what to do....

DIA [1.3K]

Answer:

my mind went... "BOOM!*

$AnimeVines$

5 0

2 years ago

To compare the average amount of time that canadians and americans spend commuting, a researcher collects a sample of canadians

yKpoI14uk [10]

The standard error of the difference of sample means is 0.444

From the complete question, we have the following parameters

Canadians

Sample size = 50
Mean = 4.6
Standard deviation = 2.9

Americans

Sample size = 60
Mean = 5.2
Standard deviation = 1.3

The standard error of a sample is the quotient of the standard deviation and the square root of the sample size.

This is represented as:

$SE = \frac{\sigma}{\sqrt n}$

The standard error of the Canadian sample is:

$SE_1 = \frac{2.9}{\sqrt{50}}$

So, we have:

$SE_1 = 0.41$

The standard error of the American sample is:

$SE_2 = \frac{1.3}{\sqrt{60}}$

So, we have:

$SE_2 = 0.17$

The standard error of the difference of sample means is then calculated as:

$SE= \sqrt{SE_1^2 + SE_2^2}$

This gives

$SE= \sqrt{0.41^2 + 0.17^2}$

$SE= \sqrt{0.197}$

Take square roots

$SE= 0.444$

Hence, the standard error of the difference of sample means is 0.444

Read more about standard errors at:

brainly.com/question/6851971

5 0

2 years ago

Can anyone thank you for me plz plz plz

serg [7]

Answer:

first what is the question

3 0

3 years ago

Read 2 more answers

The point (−2,4) lies on the curve in the xy-plane given by the equation f(x)g(y)=17−x−y, where f is a differentiable function o

Nikitich [7]

The value of $\frac{dy}{dx}$ is -3. $\blacksquare$

<h2>Procedure - Differentiability</h2><h3 /><h3>Chain rule and derivatives</h3><h3 />

We derive an expression for $\frac{dy}{dx}$ by means of chain rule and differentiation rule for a product of functions:

$\frac{d}{dx}[f(x)\cdot g(y)] = [f'(x)\cdot \frac{dx}{dx}]\cdot g(y) + f(x) \cdot [g'(y)\cdot \frac{dy}{dx} ]$

$\frac{d}{dx}[f(x)\cdot g(y)] = f'(x)\cdot g(y) +f(x)\cdot g'(y)\cdot \frac{dy}{dx}$ (1)

If we know that $f(x) \cdot g(y) = 17-x-y$ , $f(-2) = 3$ , $f'(-2) = 4$ , $g(4) = 5$ and $g'(4) = 2$ , then we have the following expression:

$-1-\frac{dy}{dx} = (4)\cdot (5) + (3)\cdot (2) \cdot \frac{dy}{dx}$

$-1-\frac{dy}{dx} = 20 + 6\cdot \frac{dy}{dx}$

$7\cdot \frac{dy}{dx} = -21$

$\frac{dy}{dx} = -3$

The value of $\frac{dy}{dx}$ is -3. $\blacksquare$

To learn more on differentiability, we kindly invite to check this verified question: brainly.com/question/24062595

<h3>Remark</h3>

The statement is incomplete and full of mistakes. Complete and corrected form is presented below:

The point (-2, 4) lies on the curve in the xy-plane given by the equation $f(x)\cdot g(y) = 17 - x\cdot y$ , where $f$ is a differentiable function of $x$ and $g$ is a differentiable function of $y$ . Selected values of $f$ , $f'$ , $g$ and $g'$ are given below: $f(-2) = 3$ , $f'(-2) = 4$ , $g(4) = 5$ , $g'(4) = 2$ .

What is the value of $\frac{dy}{dx}$ at the point $(-2, 4)$ ?

3 0

2 years ago

What is the claim? exercise and diet help people lose more weight than with the aid of this drug. This drug helps people lose mo

sdas [7]

Answer:

This drug helps people lose more weight than diet and exercise alone. The reason for this statement is that with diet and exercise alone, 23% lost weight whereas with diet, exercise and the drug, 48% lost weight so the conclusion is that the drug enhanced weight loss.

6 0

2 years ago