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4 years ago

What is the distance between the points (-4, 8) and (10,-8)?

Mathematics

1 answer:

Fiesta28 [93]4 years ago

4 0

Answer:

The distance between the 2 is (14,16)

Step-by-step explanation:

If you plot the 2 points on an online point chart and count the spaces thats how many you will find.

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What is the answer to this question?

nordsb [41]

They give you what you have to use on side of the paper and at the very bottom it said divide and now you use that and you will get your answer to your problem

8 0

4 years ago

In a study by Peter D. Hart Research Associates for the Nasdaq Stock Market, it was

GrogVix [38]

Answer:

a) 0.1108

(b) 0.0173

Step-by-step explanation:

We are given that 20% of all stock investors are retired people. A random sample of 25 stock investors is taken.

Firstly, the binomial probability is given by;

where, n = number of trails(samples) taken = 25

r = number of successes

p = probability of success and success in our question is % of

retired people i.e. 20%.

Let X = Number of people retired

(a) Probability that exactly seven are retired people = P(X = 7)

P(X = 7) =

= = 0.1108

(b) Probability that 10 or more are retired people = P(X >= 10)

P(X >= 10) = 1 - P(X <= 9)

Now, using binomial probability table, we find that P(X <= 9) is 0.98266 at n = 25, p = 0.2 and x= 9

So, P(X >= 10) = 1 - 0.98266 = 0.0173.

7 0

3 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$