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Artist 52 [7]
1 year ago
10

in a marathon race (42.195 km), the winner’s time is 2 hours, 13 minutes, and 9 seconds, while the second-place time is 2 hours,

26 minutes, and 6 seconds.shifow,removedb8d80da1d80dbeab33a3b645b3904854d9277daeadeae8d9774eb5c43aba0e6aremoved
Mathematics
1 answer:
algol [13]1 year ago
7 0

in a marathon race (42.195 km), the winner’s time is 2 hours, 13 minutes, and 9 seconds, while the second-place time is 2 hours, 26 minutes, and 6 seconds . The distance is 2.664 km.

D = distance of marathon = 42.195 km = 42195 m

tw = time taken by winner = 2 h 17 min = 2 (60) + 17 = 137 min = 137 (60) = 8220 sec

Vw = speed of winner

speed of winner is given as

Vw = D /tw = 42195/8220

ts = time taken by second place = 2 h 26 min 14 sec = 2 (60)(60) + (26)(60) + 14 = 8774 s

Vs = speed of second place

speed of second place is given as

Vs = D /ts = 42195/8774

distance of the second place holder from the finish line is given as

D' = Vs (ts - tw)

D' = (42195/8774 )(8774 - 8220)

D' = 2664.24 m

D'= 2.664 km

To learn more about distance  visit:brainly.com/question/15172156

#SPJ4

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The television show CSI: Shoboygan has been successful for many years. That show recently had a share of 18, meaning that among
jeyben [28]

Answer:

(a) The value of P (None) is 0.062.

(b) The value of P(at least one) is 0.938.

(c) The value of P(at most one) is 0.253.

(d) The event is not unusual.

Step-by-step explanation:

Let <em>X</em> = number of households watching the show.

The probability of the random variable <em>x</em> is, P (X) = <em>p</em> = 0.18.

The sample selected for the survey is of size, <em>n</em> = 14

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 14 and <em>p</em> = 0.18.

The probability of a Binomial distribution is computed using the formula:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,...

(a)

Compute the probability that none of the households are tuned to CSI: Shoboygan as follows:

P(X=0)={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}=1\times1\times0.06214=0.062

Thus, the value of P (None) is 0.062.

(b)

Compute the probability that at least one household is tuned to CSI: Shoboygan as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-0.062\\=0.938

Thus, the value of P(at least one) is 0.938.

(c)

Compute the probability that at most one household is tuned to CSI: Shoboygan as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             ={14\choose 0}(0.18)^{0}(1-0.18)^{14-0}+{14\choose 1}(0.18)^{1}(1-0.18)^{14-1}\\=0.062+0.191\\=0.253

Thus, the value of P(at most one) is 0.253.

(d)

An event that has a very low probability of occurrence is known as an unusual event.

The probability of the event "at most one household is tuned to CSI: Shoboygan" is 0.253.

This probability value is not low.

Hence, the event is not unusual.

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3 years ago
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