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s2008m [1.1K]
3 years ago
7

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

ite the function in terms of unit step function
Mathematics
1 answer:
sergij07 [2.7K]3 years ago
5 0
f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

Take it one piece at a time. For t\ge0, we can scale u(t) by -5:

-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

-5u(t-1)=\begin{cases}0&\text{for }t

so if we subtract this from -5u(t), we'll end up with

-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t

For the next piece, we can add another scaled and shifted step like

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
f(t)\equiv-5u(t)-u(t-1)+12u(t-7)
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a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

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P(success) = 64% = 0.64

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P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

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Now, we are given n = 10

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P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

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\mu = np = 92(0.64) = 58.88

\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60

We have to evaluate the probability that more than 47 cars carry just one person.

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After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

Calculation the value from standard normal z table, we have,  

P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

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