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katovenus [111]

3 years ago

15

solo el 20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal. Si

llegan 10 empleados, cuál es la probabilidad de que el guardia de seguridad encuentre

1 answer:

prisoha [69]3 years ago

3 0

Usando la distribución binomial, hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

<h3>¿Qué es la distribución binomial?</h3>

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Los parámetros son:

n es el número de ensayos.

p es la probabilidad de éxito en un ensayo

x es el número de éxitos

En este problema, hay que:

20% de los empleados de la población civil que está en una base militar restringida porta su identificación personal, o sea p = 0.2.

Llegan 10 empleados, o sea, n = 10.

La probabilidad de que el guardia de seguridad encuentre al menos uno en la base militar restringida es dada por:

$P(X \geq 1) = 1 - P(X = 0)$

En que:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

Por eso:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1074 = 0.8926$

Hay una probabilidad de 0.8926 = 89.26% de que el guardia de seguridad encuentre al menos uno en la base militar restringida.

Puede-se aprender más a cerca de la distribución binomial en brainly.com/question/25132113

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Need help on question 5 please help

blagie [28]

Answer:

$66

Step-by-step explanation:

It can be convenient to assign a different variable to the amount of money each spent. We can call the amounts spent by Seedevi, Georgia, and Amy "s", "g", and "a", respectively.

The problem statement tells us ...

s = (1/2)g

s = a +6

s + g + a = 258

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The problem statement asks for the amount Seedevi spent, so we need to find the value of s. It is convenient to write the other variables in terms of s:

g = 2s

a = s -6

Then the sum is ...

s + (2s) +(s -6) = 258

4s = 264 . . . . . . . . . . . add 6, simplify

s = 66 . . . . . . . . . . . . . .divide by 4

Seedevi spent $66.

4 0

3 years ago

A 3 foot tall globe standing next to a woman cast an 8 ft shadow. if the woman cats a shadow that is 16ft long , then how tall i

My name is Ann [436]

The woman is 6 foot tall

<h3>What is proportion?</h3>

Proportion can be defined as the mathematical comparison of two numbers.

These numbers could be numbers, elements, set or people

From the information given, we have that;

The globe is 3 feet and casts a shadow of 8ft
The woman cast a shadow of 16ft and is x feet tall

So,

if the globe is 3 foot tall and casts a shadow of 8ft

Then the woman is x foot tall to cast a shadow of 16ft

cross multiply

8x = 24

Make 'x' the subject

x = 48/ 8

x = 6 foot

Thus, the woman is 6 foot tall

Learn more about proportion here:

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2 years ago

The rent of a room is increased from rupees 2500 to rupees 3000 . Find the percent increase in rent

evablogger [386]

3,000 - 2,500 = 500

500/2,500 = 20%

The percent increase is 20%.

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3 0

3 years ago

Read 2 more answers

Vance wants to have pictures framed. Each frame and mat cost $32 and he has at most $150 to spend. Write and solve an inequality

brilliants [131]

32x< 150 , where x is the number of pictures framed. So 150/32 = 4.6875, since no pictures are purchased as whole the most he could buy is 4

5 0

3 years ago

Evaluate the following integral (Calculus 2) Please show step by step explanation!

barxatty [35]

Answer:

$\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x$

Rewrite 25 as 5²:

$\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x$

<u>Integration by substitution</u>

<u />

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=5 \sin \theta$

$\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta$

$\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta$

<u>Substitute</u> everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta$

$\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta$

$\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}$

$\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}$

$\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

$\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :$

$\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

Take out the common factor 1/2:

$\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Learn more about integration by trigonometric substitution here:

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6 0

2 years ago