Two observers collected frequency data for 10 two-minute intervals. They agreed on 8 of the intervals. What is the percentage of inter-rater reliability?

Answer:

The margin of error is 3.97 percentage points.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 600, p = 0.56[/tex]

95% confidence interval

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 - 1.96\sqrt{\frac{0.56*0.44}{600}} = 0.5203[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.56 + 1.96\sqrt{\frac{0.56*0.44}{600}} = 0.5997[/tex]

The margin of error is the upper limit subtracted by the proportion, or the proportion subtracted by the lower limit. They are the same values.

So the margin of error is 0.5997 - 0.56 = 0.56 - 0.5203 = 0.0397 = 3.97 percentage points.

Answer:

The answer to your question is

a) 400

b) The function has two real solutions (-2 and 8)

Step-by-step explanation:

Process

1.- Discriminant = b² - 4ac

                         = 12² -4(-2)(32)

                         = 144 + 256

                        = 400

2.- Solutions (using the general formula)

                   x = [tex]\frac{- b +/- \sqrt{b^{2}- 4ac}}{2a}[/tex]

                   x = [tex]\frac{- 12 +/- \sqrt{400}}{2(-2)}[/tex]

                   x = [tex]\frac{-12 +/- 20}{2(-2)}[/tex]

   x₁ = [tex]\frac{-12 + 20}{2(-2)} = \frac{-12 + 20}{-4} = \frac{-8}{4} = - 2[/tex]

   x₂ = [tex]\frac{- 12 - 20}{- 4} = \frac{-32}{- 4} = 8[/tex]

This function has two real solutions (-2, 8)

Answer: C- The value of the discrimination is 400.

E- The function has 2 real solutions.

Step-by-step explanation: Hope that helped!

Answer:  The required laplace transform of g(t) is [tex]\dfrac{5}{(s+5)^2}.[/tex]

Step-by-step explanation:  We are given to find the laplace transform of the following function :

[tex]g(t)=5te^{-5t}.[/tex]

We know the following formulas for laplace transform :

[tex](i)~L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}},\\\\(ii)~L\{cf(t)\}=cL\{f(t)\}.[/tex]

In the given function function, we have

c = 5,  n = 1  and  a = -5.

Therefore, we get

[tex]L\{g(t)\}\\\\=L\{5te^{-5t}\}\\\\=5L\{te^{-5t}\}\\\\\\=5\times\dfrac{1!}{(s-(-5))^{1+1}}\\\\\\=\dfrac{5}{(s+5)^2}.[/tex]

Thus, the required laplace transform of g is [tex]\dfrac{5}{(s+5)^2}.[/tex]

The Laplace transform of the given function is [tex]\frac{5} { (s + 5)^2}[/tex].

The Laplace transform of a function g(t) is defined as:

[tex]L{(g(t))} = \int\limits^{\infty}_0 e^-^s^tg(t) dt[/tex]

We need to find the Laplace transform of [tex]g(t) = 5te^-^5^t[/tex]. To do this, we use the shifting theorem and known transforms.

First, recall the Laplace transform of [tex]nte^-^a^t[/tex] is:

[tex]L(nte^-^a^t)} = \frac{n! } {(s + a)^n^+^1}[/tex]

For our function [tex]g(t)[/tex] :

a = 5

n = 1

Applying the formula:

[tex]L{(5te^-^5^t)} = \frac{(5 * 1! )}{(s + 5)^2}[/tex]

After simplifying, we get:

[tex]L\leftparanthesis(\ 5te^-^5^t)\rightparanthesis\ = \frac{5 }{ (s + 5)^2}[/tex]

Answer:

11)y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]

12)y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]

13)y = [tex]5e^{-(x+1)}[/tex]

14)y = 0

Step-by-step explanation:

Given data:

[tex]y=c_{1} e^{x} +c_{2} e^{-x}[/tex]

y''-y=0

The equation is

[tex]m^{r}[/tex]-1 = 0

(m-1)(m+1) = 0

if  above equation is zero then either

m - 1 = 0 or  m + 1 = 0

m = 1        ,    m  = - 1

11)

y(0) = 1 , y'(0) = 2

[tex]y'=c_{1} e^{x} -c_{2} e^{-x}[/tex]

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] = 1   (y(0) = 1) (1)

[tex]c_{1}[/tex] -  [tex]c_{2}[/tex] = 2   (y'(0) = 2)  (2)

adding 1 & 2

2[tex]c_{1}[/tex] = 3

[tex]c_{1}[/tex] = 3/2

3/2 +  [tex]c_{2}[/tex] = 1

[tex]c_{2}[/tex]  = 1 -  3/2

[tex]c_{2}[/tex] = - 1/2

y = [tex]\frac{3}{2} e^{x} - \frac{1}{2} e^{-x}[/tex]

12)

y(0) = 1 , y'(0) = e

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] = 0 (y(0) = 1) (3)

[tex]c_{1}[/tex] = - [tex]c_{2}[/tex]

[tex]e=c_{1} e -c_{2} e^{-1}[/tex]   (y'(0) = 2)  (4)

[tex]e=c_{1} e -\frac{c_{2} }{e} }[/tex]

[tex]e =\frac{c_{1} e^{2} -c_{2} }{e} }[/tex]

[tex]e^{2} ={c_{1} e^{2} -c_{2} }[/tex]

replace [tex]c_{2}[/tex] = [tex]c_{1}[/tex] by equation 3

[tex]e^{2} ={c_{1} e^{2} -c_{1} }[/tex]

taking common [tex]c_{1}[/tex]

[tex]e^{2} =c_{1} ({e^{2} -1 })[/tex]

[tex]\frac{e^{2} }{({e^{2} -1 })} =c_{1}[/tex]

[tex]-\frac{e^{2} }{({e^{2} -1 })} =c_{2}[/tex]

∴ y = [tex]\frac{e^{2} }{1+e^{2} } (e^{x} - e^{-x} )[/tex]

13)

y(-1) = 5 , y'(-1) = -5

[tex]c_{1}[/tex][tex]e^{-1}[/tex] +  [tex]c_{2}[/tex][tex]e^{1}[/tex] = 5   (y(-1) = 5 ) (5)

[tex]c_{1}[/tex][tex]e^{-1}[/tex] -  [tex]c_{2}[/tex][tex]e^{1}[/tex] = -5    (y'(-1) = -5)  (6)

Adding 5&6

2[tex]c_{1}[/tex] [tex]e^{-1}[/tex] = 0

[tex]c_{1}[/tex] = 0

[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - [tex]c_{1}[/tex][tex]e^{-1}[/tex]

[tex]c_{2}[/tex][tex]e^{1}[/tex] = 5 - 0

[tex]c_{2}[/tex]= 5/e

y = [tex]5e^{-1} e^{-x}[/tex]

y = [tex]5e^{-(x+1)}[/tex]

14)

y(0) = 0 , y'(0) = 0

[tex]c_{1}[/tex] +  [tex]c_{2}[/tex] =  0 (y(0) = 0) (7)

[tex]c_{1}[/tex] -  [tex]c_{2}[/tex] = 0   (y'(0) = 0)  (8)

Adding 7 & 8

2[tex]c_{1}[/tex] = 0

[tex]c_{2}[/tex] =

y = 0

The tangent line approximation near x=0 for the function f(x) = \\sqrt{10 + x} is found by first calculating its derivative, then using that derivative to construct the equation of the tangent line at x=0, resulting in the linear approximation y = (1/2)(10)^{-1/2}x + \\sqrt{10}.

Finding the tangent line approximation for a function near a point involves using the function's derivative at that point. For the function f(x) = \\sqrt{10 + x}, the derivative at x = 0, denoted as f'(0), will provide the slope of the tangent. To find this, let's differentiate f(x) using the chain rule. The derivative of f(x) with respect to x is (1/2)(10 + x)^{-1/2}. At x = 0, this simplifies to 1/2(\\sqrt{10}), which is the slope of the tangent line at that point. Hence, the tangent line equation is y - f(0) = f'(0)(x - 0), which simplifies to y = (1/2)(10)^{-1/2}x + \\sqrt{10}. This form equation is the linear approximation of f(x) near x = 0.

Answer:

P = 4s

Step-by-step explanation:

The perimeter of a geometric shape is simply the sum of all its sides length. Since the shape in question is a square, which means that all of the four sides have the same length 's', the perimeter can be expressed by:

[tex]P = s+s+s+s\\P=4s[/tex]

For any value of 's', the formula above expresses the perimeter 'P' as a function of 's'

Answer:

=19

Step-by-step explanation:

=6÷3+17

=2+17

=19

Answer:

a) [tex]P(X=12)=(12C12)(0.2)^{12} (1-0.2)^{12-12}=4.096x10^{-9}[/tex]

b) [tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313 [/tex]

c) [tex] E(X) = np = 12*0.2= 2.4[/tex]

d) [tex] Sd(X) = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386[/tex]

e) [tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558[/tex]

f)  [tex] P(X\leq 10) =1.1x10^{-9}[/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=12, p=0.2)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Part a

For this case we want to find this probability:

[tex]P(X=12)=(12C12)(0.2)^{12} (1-0.2)^{12-12}=4.096x10^{-9}[/tex]

Part b

[tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0) [/tex]

[tex]P(X=0)=(12C0)(0.2)^{0} (1-0.2)^{12-0}=0.0687[/tex]

[tex]P(X \geq 1) =1-P(X<1)= 1-P(X=0)=1-0.0687=0.9313 [/tex]

Part c

The expected value is given by:

[tex] E(X) = np = 12*0.2= 2.4[/tex]

Part d

The standard deviation is given by:

[tex] Sd(X) = \sqrt{np(1-p)}=\sqrt{12*0.2*(1-0.2)}=1.386[/tex]

Part e

If we want the probability that the number classified as deceptive would be lower than the mean we want:

[tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)[/tex]

[tex]P(X=0)=(12C0)(0.2)^{0} (1-0.2)^{12-0}=0.0687[/tex]

[tex]P(X=1)=(12C1)(0.2)^{1} (1-0.2)^{12-1}=0.2062[/tex]

[tex]P(X=2)=(12C2)(0.2)^{2} (1-0.2)^{12-2}=0.2835[/tex]

[tex] P(X<2.4) =P(X\leq2) =P(X=0) +P(X=1)+P(X=2)= 0.558[/tex]

Part f

For this case our random variable would be:

[tex]X \sim Binom(n=200, p=0.2)[/tex]

And we want this probability:

[tex] P(X\leq 10) = P(X=0)+P(X=1)+ .......+P(X=10)[/tex]

And we can use the following excel code to find the answer:

"=BINOM.DIST(10;200;0.2;TRUE)"

And we got: [tex] P(X\leq 10) =1.1x10^{-9}[/tex]

Answer:

The length of the circular arc is 8.64 inches

Step-by-step explanation:

Length of circular arc (L) = central angle/360° × 2πr

central angle = pie/4 = 45°, r (radius) = 11 inches

L = 45°/360° × 2 × 3.142 × 11 = 8.64 inches (to two decimal places)

Answer:

Step-by-step explanation:

The formula for determining the length of an arc is expressed as

Length of arc = θ/360 × 2πr

Where

θ represents the central angle.

r represents the radius of the circle.

π is a constant whose value is 3.14

From the information given,

Radius, r = 11 inches

θ = pi/4

2π = 360 degrees

π = 360/2 = 180

Therefore,

θ = 180/4 = 45 degrees

Therefore,

Length of arc = 45/360 × 2 × 3.14 × 11

Length of arc = 8.64 inches rounded up to 2 decimal places

Answer:

the explanation is given below.

Step-by-step explanation:

from this, the lowest number is 1 and the highest number is 100

Answer:

The probability that at most 3 attempts are required to produce a craft item with satisfactory quality is 0.9375

Step-by-step explanation:

Let E be a random variable denoting the event that an artist makes a craft item with satisfactory level.

Then the random variable E follows a Geometric distribution.

A Geometric distribution is defined as the number of failures (k) before the first success.

The probability function of Geometric distribution is:

[tex]P(X=k)=(1-p)^{k}p[/tex], p = Probability of success and k = 0, 1, 2, 3...

The probability of success is, p = 0.5 and the number of failures is, k = 3.

Compute the probability of at most 3 attempts before the first success is:

[tex]P(X\leq 3) =P(X=3)+P(X = 2)+P(X=1) +P(X = 0)\\=[(1-0.5)^{0}*0.5]+[(1-0.5)^{1}*0.5]+[(1-0.5)^{2}*0.5]+[(1-0.5)^{3}*0.5]\\=0.9375[/tex]

Therefore, the probability that at most 3 attempts are required to produce a craft item with satisfactory quality is 0.9375.

To design a sine function with a period of 24 and minima and maxima at given times, we scale the function using a coefficient of 2π/24, shift the function to make the peak occur at x=16, and stretch it by a factor of 3 to make it go from 10 to 16.

To design a sine function with a period of 24 and minima and maxima at given times, we first need to understand a few concepts about sine functions. The standard sine function, sine(x), has a period of 2π. Therefore, to stretch it to a period of 24 hours, we would scale the function using a coefficient of 2π/24 or π/12. Thus, our function becomes sine((π/12) x).

Next, we want to shift the function so its maximum occurs at t=16. Normally, the sine function peaks at π/2, so we need to shift the function to the right by an amount that makes the peak occur at x=16. This would be 16 - π/2, which gives us the function sine(π/12 x - 16 + π/2).

Finally, to stretch the function vertically to accommodate the minimum and maximum values of 10 and 16, we note that the amplitude of the sine function is usually 1 (from -1 to 1), so we need to stretch it by a factor of (16-10)/2 = 3 to make it go from 10 to 16. This gives us the function y= 3sin((π/12)x - 16 + π/2)+13.

https://brainly.com/question/32247762

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Answer:

im pretty sure you have to find out how far Joel walked and then double joels distance to get Brents

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

a. 24ways

b.256ways

Step-by-step explanation:

the letters IOWA contains for letters, since we are to arrange without repeating any letter, we permutate the letters.

For permutation of n object in r ways is expressed as

P(n,r)=n!/(n-r)!

hence for n=4 and r=4, we have P(4,4)=4!/(4-4)!

P(4,4)=4!/(0)!

P(4,4)=4*3*2*1=24ways

b. To arrange the letters such that each letter can be repeated, we can arrange the letter I in four ways, letter O can be arrange in four ways, letter W can be arranged in four ways and letter A can be arranged in four ways ..

Hence we arrive at

4*4*4*4=256ways

Answer:

8.99 days elapsed. Option (C) is correct

Step-by-step explanation:

The distribution  has density function k/t+1 for a constant k and t between 0 and 6.5 . Since the distribution is symmetrical in 6.5, the area it forms between 0 and 6.5 should be 1/2, thus

[tex]\frac{1}{2} = \int\limits_0^{6.5} \frac{k}{t+1} \, dt = k *(ln(t+1) \, |_0^{6.5}) = k * (ln(7.5)-ln(1)) = k*ln(7.5)[/tex]

Hence k = 1/(2ln(7.5)), approx 1/4.

We need to find the percentil 0.6, since the integral of the random variable is 1/2 over the first half, we need to find t such that the integral of the random variable between o and 6.5 + t is 0.6. This is equivalent to find t such that the integral between 6.5 and 6.5+t is 0.1. Due to the  over 6.5, this t should satisfy that the integral between 6.5-t and 6.5 is also 0.1. Lets compute the integral and find t

[tex]\int\limits^{6.5}_{6.5-t} {\frac{k}{t+1}} \, dx = \frac{1}{2ln(7.5)}*(ln(t+1) \, |_{6.5-t}^{6.5} \, ) = \frac{1}{2ln(7.5)} * (ln(7.5)-ln(7.5-t)) = \\\frac{1}{2} - \frac{ln(7.5-t)}{2ln(7.5)} = 0.1[/tex]

Therefore,

[tex]\frac{ln(7.5-t)}{2ln(7.5)} = 0.4\\\\ln(7.5-t) = 0.8*ln(7.5)\\\\7.5-t = e^{0.8*ln(7.5)}\\\\t = 7.5-e^{0.8*ln(7.5)} = 2.49[/tex]

As a result, the student sould have registered 2.49 days after the day 6.5, thus it should have registeredd at day 8.99. Option (C) is correct.

Answer: ( Min = 0 , [tex]Q_1=1[/tex] , [tex]Q_2=3[/tex] , [tex]Q_3=7[/tex] , Max = 10 )

Step-by-step explanation:

Given : A random sample of 11 days were selected from last year's records maintained by the maternity ward in a local hospital, and the number of babies born each day of the days is given below:

3  7   7   10    0    7    1    2    5   3    0

We first arrange them in increasing order , we get

0    0    1    2   3   3   5     7   7   7   10

Here , N= 11

Now , we can see that

Minimum value = 0

Maximum value = 10

First quartile [tex]Q_1[/tex]= [tex](\dfrac{N+1}{4})^{th}\ term=(\dfrac{12}{4})^{th}\ term = 3^{rd} term =1[/tex]

Second quartile [tex]Q_2[/tex]= Median = Middlemost number = 3

Third quartile [tex]Q_3[/tex] =  [tex](\dfrac{3(N+1)}{4})^{th}\ term=(\dfrac{36}{4})^{th}\ term[/tex]

[tex]= 9^{th} term =7[/tex]

∴ The required five number summary : ( Min = 0 , [tex]Q_1=1[/tex] , [tex]Q_2=3[/tex] , [tex]Q_3=7[/tex] , Max = 10 )

Answer: individual

Step-by-step explanation:

An individual is a person or object that is a member of the population being studied. A population is defined as a group of individuals with a common characteristic living and interbreeding within a given area, in statistics, population is a collection of individuals to be studied. Individuals can also be referred to as the objects/person described by a set of data. For example: when studying the height of students in a school, the students attending that school are individuals.

Answer:

6 is a factor of n

Step-by-step explanation:

2 is a factor of n and 3 is a factor of n means

n = 2×3×k

  = 6×k

then  n = 6×k

then 6 is a factor of n

Final answer:

If both 2 and 3 are factors of a number n, then 6 must also be a factor of n, because the product of unique prime factors is always a factor of that number.

Explanation:

The product of unique prime factors of a number will be a factor of that number. Since 2 and 3 are prime factors and both are factors of n, their product (which is 6) must also be a factor of n.

For example, consider the number 12. 12 is divisible by 2 and 12 is divisible by 3, and indeed, 12 is divisible by 6 as well. This holds true for any number n that has 2 and 3 as factors. Thus, we can conclude that 6 is a factor of n if both 2 and 3 are factors of n.

Answer:  The proofs are given below.

Step-by-step explanation:  We are given to prove that the following statements are tautologies using truth table :

(a) ¬r ∨ (¬r → p)                              b. ¬(p → q) → ¬q

We know that a statement is a TAUTOLOGY is its value is always TRUE.

(a) The truth table is as follows :

r                 p                 ¬r                       ¬r→p                     ¬r ∨ (¬r → p)

T                T                   F                         T                                T

T                F                   F                         T                                T

F                T                   T                         T                                T

F                F                   T                         F                                T  

So, the statement (a) is a  tautology.

(b) The truth table is as follows :

p                 q                 ¬q                       p→q             ¬(p→q)          ¬(p→q)→q

T                T                   F                         T                      F                    T

T                F                   T                         F                      T                    T

F                T                   F                         T                      F                    T

F                F                   T                         T                       F                   T

So, the statement (B) is a  tautology.              

Hence proved.

Final answer:

To prove that a statement is a tautology using propositional logic, we need to show that the statement is true under all possible truth values of its variables. By applying the laws of implication, disjunction, and contradiction, we can prove that the given statements are tautologies.

Explanation:

To prove that a statement is a tautology using the laws of propositional logic, we need to show that the statement is true under all possible truth value assignments of its variables. Let's consider each statement:

a. ¬r ∨ (¬r → p)

We can use the law of implication, which states that ¬p ∨ q is equivalent to p → q, to rewrite the statement as ¬r ∨ (r → p). By applying the law of disjunction, which states that p ∨ (q ∧ r) is equivalent to (p ∨ q) ∧ (p ∨ r), we can further rewrite the statement as (¬r ∧ r) ∨ (r ∨ p). Using the law of contradiction, which states that p ∧ ¬p is always false, we can simplify the statement to r ∨ p, which is a tautology.

b. ¬(p → q) → ¬q

We can use the law of implication to rewrite the statement as ¬(¬p ∨ q) → ¬q. By applying De Morgan's law, which states that ¬(p ∨ q) is equivalent to ¬p ∧ ¬q, we can simplify the statement to (p ∧ ¬q) → ¬q. Using the law of contradiction, we know that p ∧ ¬p is always false, so the statement simplifies to false → ¬q, which is always true. Therefore, it is a tautology.

Answer:

Option C: The line y = x

Step-by-step explanation:

Let us take an ordered pair (x, y) of a function. Then the ordered pair of its inverse function would be (y, x).

That is to say, when we reflect a point (x, y) across the line y = x we get the point (y, x).

Note that since, this function is invertible, it is both 'one - one' and 'onto'.

The line of reflection for two functions with an inverse relationship is the line y = x, as this line shows the symmetry of the original function and its inverse on a graph.

When two functions have an inverse relationship, the line of reflection across which their graphs are mirrored is the line y = x. This is because for an inverse relationship, each x value in the first function corresponds to a y value in the second function, and vice versa. Thus, when graphing the original function and its inverse, you will notice that they are symmetric with respect to the line y = x.

Answer:

yes

Step-by-step explanation:

the rate of change is constant

Answer:  0.1

Step-by-step explanation:

WE know that the total probability in an experiments = 1.

i.e. Sum of the probabilities of occurring each event is 1.

i.e. If there are n outcomes in any experiment., then the total probability will be:

[tex]P(E_1)+P(E_2)+P(E_3)+...........+P(E_n)=1[/tex]

Given : An experiment consists of four outcomes ,with P(E1) = 0.2, P(E2) = 0.3, and P(E3) = 0.4.

Then , [tex]P(E_1)+P(E_2)+P(E_3)+P(E_4)=1[/tex]

Substitute corresponding values , we get

[tex]0.2+0.3+0.4+P(E_4)=1[/tex]

[tex]0.9+P(E_4)=1[/tex]

[tex]P(E_4)=1-0.9=0.1[/tex]

Hence , the probability of outcome [tex]E_4[/tex] is 0.1.

The probability of event E4 in the given experiment is 0.1, because the sum of probabilities of all outcomes should equal 1.

The problem falls under the subject of

Probability Theory

within Mathematics. In probability, the total probability of all possible outcomes is always 1. So, for the given problem where you have four events E1, E2, E3, E4, the total probability P(E1)+P(E2)+P(E3)+P(E4) should equal 1. Given that P(E1) = 0.2, P(E2) = 0.3 and P(E3) = 0.4, we can find P(E4) using the equation

1 - P(E1) - P(E2) - P(E3) = P(E4)

. By substituting the given values into this equation, we find that P(E4) = 1 - 0.2 - 0.3 - 0.4 =

0.1

. Therefore, the probability of outcome E4 is 0.1.

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Answer:

20

Step-by-step explanation:

You could find 5/⅓

by using 5 × 3

Knowing that:

i. Any number multiplied by 1, gives the number itself.

ii. Dividing any number by itself gives 1.

You would agree with me that

i. (5×3)/(5×3) = 1

ii. Writing 5/⅓ as 5/⅓ × 1 doesn't change the value.

Then I can write 5/⅓ as

5/⅓ × (5×3)/(5×3) = 1

This can become

[5×(5×3)] / [(⅓) × (5×3)]

= 75/(15/3)

= 75/5

= 15

In a similar way,

12/ 3/5

= [12/ (3/5)] × [(5×3)/(5×3)]

= 12×(5×3) / (3/5)×(5×3)

= (12×5×3) / [(3×5×3)/5]

= 180 / (45/5)

= 180 / 9

= 20

The probability that a single randomly selected salary is at least $134,000 is approximately 0.5120 or 51.20%.

To find the probability that a single randomly selected salary is at least $134,000, we need to calculate the z-score and use the standard normal distribution.

1. Calculate the z-score:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case, x = $134,000, μ = $133,000, and σ = $31,000.

z = (134000 - 133000) / 31000

z = 1 / 31000

z ≈ 0.0323

2. Find the probability associated with the z-score:

We can use a z-table or a calculator to find the probability.

From the z-table, we find that the probability corresponding to a z-score of 0.0323 is approximately 0.5120.

Therefore, the probability that a single randomly selected salary is at least $134,000 is 0.5120 or 51.20%.

Answer:

Step-by-step explanation:

The probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5 is 6/19,

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

It is given that:

The randomly chosen number between 1 and 100 is divisible by 3

Applying conditional probability:

Let A is the event: the numbers divisible by 3

Let B is the event: At least one digit equal to 5

P(A|B) = n(A∩B)/n(B)

P(A|B) = 6/19

Thus, the probability that a randomly chosen number between 1 and 100 is divisible by 3, given that the number has at least one digit equal to 5 is 6/19.

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Answer:

1) The solution of the system is

[tex]\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right[/tex]

2) The solution of the system is

[tex]\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right[/tex]

Step-by-step explanation:

1) To solve the system of equations

[tex]\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right[/tex]

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

[tex]\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]

Step 2: Swap rows 1 and 2

[tex]\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]

Step 3:  [tex]\left(R_1=\frac{R_1}{6}\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right][/tex]

Step 4: [tex]\left(R_3=R_3-R_1\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right][/tex]

Step 5: [tex]\left(R_2=\frac{R_2}{3}\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right][/tex]

Step 6: [tex]\left(R_3=R_3+R_2\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right][/tex]

Step 7: [tex]\left(R_3=\left(\frac{6}{37}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]

Step 8: [tex]\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]

Step 9: [tex]\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right][/tex]

Step 10: Rewrite the system using the row reduced matrix:

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right[/tex]

2) To solve the system of equations

[tex]\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right[/tex]

using the row reduction method you must:

Step 1:

[tex]\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]

Step 2: [tex]\left(R_1=\frac{R_1}{4}\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]

Step 3: [tex]\left(R_2=R_2-\left(2\right)R_1\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right][/tex]

Step 4: [tex]\left(R_3=R_3-R_1\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]

Step 5: [tex]\left(R_2=\left(2\right)R_2\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]

Step 6: [tex]\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right][/tex]

Step 7: [tex]\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)[/tex]

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right][/tex]

Step 8: [tex]\left(R_3=\left(- \frac{2}{117}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]

Step 9: [tex]\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]

Step 10: [tex]\left(R_2=R_2-\left(15\right)R_3\right)[/tex]

[tex]\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right][/tex]

Step 11:

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right[/tex]

Answer:

A.

sigma > 22500

B.

Null hypothesis:sigma = 22500

Alternative hypothesis:sigma > 22500

Step-by-step explanation:

A.

The claim states that the standard deviation of high school teachers income  is more than 22,500. This can be represented in the symbolic form as sigma > 22500.

B.

The null hypothesis and alternative hypothesis for the given scenario can be written as

Null hypothesis: Standard deviation of income of high school teachers is 22,500.

The standard deviation is represented as sigma.Symbolically it can be written as

Null hypothesis: sigma = 22500

Alternative hypothesis: Standard deviation of income of high school teachers is more than 22,500.

Symbolically it can be written as

Alternative hypothesis: sigma > 22500

[tex]x''+4x'+53x=15[/tex]

has characteristic equation

[tex]r^2+4r+53=0[/tex]

with roots at [tex]r=-2\pm7i[/tex]. Then the characteristic solution is

[tex]x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)[/tex]

For the particular solution, consider the ansatz [tex]x_p=a_0[/tex], whose first and second derivatives vanish. Substitute [tex]x_p[/tex] and its derivatives into the equation:

[tex]53a_0=15\implies a_0=\dfrac{15}{53}[/tex]

Then the general solution to the equation is

[tex]x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}[/tex]

With [tex]x(0)=8[/tex], we have

[tex]8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}[/tex]

and with [tex]x'(0)=-19[/tex],

[tex]-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}[/tex]

Then the particular solution to the equation is

[tex]\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}[/tex]

The given equation is a second order linear differential equation. However, there seems to be an inconsistency with the constant right-hand side. In a well-formulated equation, one would solve this by using characteristic roots and trivial solutions, finding the general solution, then applying initial conditions.

The given equation is a second order linear differential equation. Given the initial conditions, including the fact that x(0) = 8 and x˙ = −19, one must adjust for these when solving the differential equation. By utilizing the characteristic equation for determining the roots, the solution and consequent constants are then determined.

However, please note that without some form of driving force (right-hand side function), this is a simple harmonic oscillator equation. Since the right hand side function (15 in this case) is constant, there's an inconsistency in the problem. In this context, for a correct form, it should have a time-dependent function on the right side. If we assume that an inconsistency has occurred and the right side is zero, a full solution could be given.

In a correct equation scenario, one would be able to solve the initial value problem with characteristic roots and trivial solutions, find the particular solution, then a general solution and apply initial value conditions to find specific constants.

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Answer:

No, because it is not the tens digit that goes up by 1 in these numbers, it is the unit digit.

Step-by-step explanation:

It is important to know the concepts of units, tenths and cents.

For example

1 = 1 unit

10 = 1*10 + 0 = The tens digit is one the unit digit is 0

21 = 2*10 + 1 = The tens digit is two and the unit digit is 1.

120 = 1*100 + 2*10 + 0 = The cents digit is 1, the tens digit is two and the unit digit is 0.

So

Adding 1 is the same as the unit digit going up by 1.

Adding 10 is the same as the tens digit going up by 1.

Adding 100 is the same as the cents digit going up by 1.

In this problem, we have that:

864,865,866,867,868,869

Each value is the 1 added to the previous value, that is, the unit digit goes up by 1.

Mariel thinks the tens digit goes up by 1 in these numbers. Do you agree?

No, because it is not the tens digit that goes up by 1 in these numbers, it is the unit digit.

Answer:

disagree, its the unit number that goes up by 1

Step-by-step explanation:

Final answer:

Using the pigeonhole principle, we can prove that among 502 positive integers, at least two will have the same remainder when divided by 1000, implying their difference is divisible by 1000.

Explanation:

The question asks to prove that among 502 positive integers, there are always two integers so that either their sum or their difference is divisible by 1000. This statement can be understood through the pigeonhole principle, which in basic terms means if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon.

In this case, consider the remainders when these integers are divided by 1000. Since there are only 1000 possible remainders (from 0 to 999), and we have 502 numbers, at least two of them must have the same remainder when divided by 1000, according to the pigeonhole principle.

Let these two numbers be a and b, where without loss of generality, a ≥ b. If a and b have the same remainder when divided by 1000, then a - b is divisible by 1000. Alternatively, if we had a case where the sum is considered, assuming complementary pairs mod 1000, a similar argument involving the pigeonhole principle can conclude that there must be at least one pair whose sum or difference gives a number divisible by 1000, satisfying the initial claim.