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Dafna11 [192]
2 years ago
9

At 111111:55\text{ p.M.}55 p.M.55, start text, space, p, point, m, point, end text, Thomas ties a weight to the minute hand of a

clock. The clockwise torque applied by the weight (i.E. The force it applies on the clock's hand to move clockwise) varies in a periodic way that can be modeled by a trigonometric function. The torque peaks 151515 minutes after each whole hour, when the minute hand is pointing directly to the right, at 3\text{ Nm}3 Nm3, start text, space, N, m, end text (Newton metre, the SI unit for torque). The minimum torque of -3\text{ Nm}−3 Nmminus, 3, start text, space, N, m, end text occurs 151515 minutes before each whole hour, when the minute hand is pointing directly to the left.
Mathematics
1 answer:
Dahasolnce [82]2 years ago
7 0

Answer:

r(t)=3sin((2pi/60)*(t-5))

Step-by-step explanation:

Found it on Khan Academy

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The manufacturer of an electronic device claims that the probability of the device failing during the warranty period is 0.005.
Zielflug [23.3K]

Answer:

The  value is  P(X \ge  15) =  0.5

Step-by-step explanation:

From the question we are told that

    The probability of the device failing during the warranty period is p =  0.005

    The  sample size is  n = 3000

     The random variable  considered is  x  =  15

Generally this is distribution is binomial given the fact that there is only two out comes hence

      X  which is a variable representing a randomly selected selected electronic follows a binomial distribution i.e

     X  \~ \  B(n , p)

Now the mean is mathematically evaluated as

      \mu  =  n *  p

=>   \mu  = 3000 *  0.005

=>    \mu  =15

The standard deviation is mathematically represented as

      \sigma  =  \sqrt{np(1 -p )}

=>  \sigma  =  \sqrt{3000 *  0.005 * (1 - 0.005 )}

=>   \sigma  =  3.86

Now given that n is very large, then it mean that we can successfully apply normal approximation on this  binomial distribution

So

    P(X \ge  15) =  P( \frac{X - \mu}{\sigma }  \ge \frac{x - \mu}{\sigma } )

Now  applying  Continuity Correction we have

   P(X \ge  (15-0.5)) =  P( \frac{X - \mu}{\sigma }  > \frac{(15 -0.5) - 15}{3.86 } )

Generally  \frac{X - \mu}{\sigma }   =  Z (The \ standardized \ value  \ of  \ X)

    P(X \ge  (15-0.5)) =  P(Z >-0.130 )

From the z-table  

       P(Z >-0.130 )  =0.5

Thus  

    P(X \ge  15) =  0.5

 

   

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Which of the following is a solution to the equation c+(4-3c)-2=0
Eduardwww [97]

Answer:

c=1

Step-by-step explanation:

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