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2 years ago

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Statisticians for a roadside assistance company interviewed 50,000 randomly selected United States (US) households. Of those, 15

,595 reported that they had traveled 50 or more miles from home between December 23 and January 4. If there are 115,000,000 US households, approximately how many of them do the interviews suggest traveled 50 or more miles from home at that time?

1 answer:

I am Lyosha [343]2 years ago

8 0

Answer:

35,868,500 people

Step-by-step explanation:

#We calculate the probability of those who traveled 50+ miles.

-Probability is the likelihood of success. n=50000, p=15595:

$P(X\geq 50)=\frac{15595}{50000}\\\\=0.3119$

-The US population is 115000000. We determine what fraction of the actual population traveled 50+ miles:

$E(X)=np, N=115000000,p=0.3119\\\\=0.3119\times 115000000\\\\\\=35868500$

Hence, 35,868,500 people traveled 50+ miles.

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

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2 years ago

A row of tiny red beads is placed at the beginning and at the end of a 20‐centimeter bookmark.

AlladinOne [14]

Answer:

51 rows

Step-by-step explanation:

Given

Length of bookmark = 20cm

Distance between beads = 4mm

Required

Number of rows of beads

First, the distance between the rows of beads must be converted to cm

if 1mm = 0.1cm

then

4mm = 4*0.1cm

4mm = 0.4 cm

This means that each row of beads is placed at 0.4 cm mark.

The distance between each row follows an arithmetic progression and it can be solved as follows;

$T_n = a + (n-1)d$

Where $Tn = 20cm$ (The last term)

$a = 0 cm$ (The first term)

$d = 0.4cm$ (The distance between each row of beads)

n = ?? (number of rows)

Solving for n; we have the following;

$T_n = a + (n-1)d$ becomes

$20 = 0 + (n-1)0.4$

$20 = (n-1)0.4$

DIvide both sides by 0.4

$\frac{20}{0.4} = \frac{(n-1)0.4}{0.4}$

$50 = (n-1)$

$50 = n-1$

Add 1 to both sides

$50 + 1= n-1 + 1$

$n = 51$

Hence, the number of rows of beads is 51

4 0

2 years ago

A gym charges a $20 sign-up fee plus $25 per month. You have a $120 gift card for the gym. When does the total spent on your gym

kati45 [8]

After 4 months! You do 120-20=100 which indicates only the month prices, then 100 divided by 25 which is 4!

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2 years ago

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as L = 10 lo

butalik [34]

Answer:

The approximate loudness of a rock concert with a sound intensity of $10^{-1}$ is 110 Db.

Step-by-step explanation:

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is : $L=10 log (\frac{I}{I_0})$

$I_0=10^{-12}$

We are supposed to find What is the approximate loudness of a rock concert with a sound intensity of $10^{-1}$

So, $I = 10^{-1}$

Substitute the values in the formula :

$L=10 log (\frac{10^{-1}}{10^{-12}})$

$L=10 log (10^{11})$

L=110 Db

So, the approximate loudness of a rock concert with a sound intensity of $10^{-1}$ is 110 Db.

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VikaD [51]

Answer:

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Step-by-step explanation:

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