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juin [17]
3 years ago
9

The video states that 78% of people carry less than $50 in cash; 40% carry less than $20 in cash, and 9% carry no cash. Suppose

you have a random sample of 100 people and you check how much cash each person is carrying. Assuming that the percentages stated in the video are correct for all adult Americans, discuss how it could occur that 12 people in your sample are carrying no cash while only 75 people in your sample are carrying less than $50 in cash.
Mathematics
1 answer:
Inessa05 [86]3 years ago
6 0

Answer:

The probability that 12 people in your sample are carrying no cash is 0.0712

Step-by-step explanation:

n = 100

p(no cash) = 0.09

x = 12

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 12) = 0.074.

The probability that 12 people in your sample are carrying no cash is 0.074.

n = 100

p(less than 50) = 0.78

x = 75

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 75) = 0.0712

The probability that 12 people in your sample are carrying no cash is 0.0712

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Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
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Answer:

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Step-by-step explanation:

According to Taylor's theorem

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for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

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In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

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\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

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