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irakobra [83]
2 years ago
8

60% of 145 is what number.

Mathematics
2 answers:
mina [271]2 years ago
8 0
 the answer is 87   60% or 145 = 87
Hoochie [10]2 years ago
8 0
Answer: 261

To find the answer follow these steps:
1: To find the number or 100% find 10% of 145 (14.5) and subtract it from 145 (you will get 130.5.
  - We do this because going down 10% you can get to 50% of the number then easily multiply by 2.
2: Now multiply 130.5 by 2 and you will get 261.

Hope this helps!
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Let X be the waiting time for a car to pass by on a country road, where X has an average value of 35 minutes. If the random vari
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Answer:

Probability that the wait time is greater than 37 minutes is 0.3474.

Step-by-step explanation:

We are given that the random variable X is known to be exponentially distributed and X be the waiting time for a car to pass by on a country road, where X has an average value of 35 minutes.

<u><em>Let X = waiting time for a car to pass by on a country road</em></u>

The probability distribution function of exponential distribution is given by;

f(x) = \lambda e^{-\lambda x}  , x >0     where, \lambda = parameter of distribution.

Now, the mean of exponential distribution is = \frac{1}{\lambda}  which is given to us as 35 minutes that means  \lambda = \frac{1}{35}  .

So, X ~ Exp( \lambda = \frac{1}{35} )

Also, we know that Cumulative distribution function (CDF) of Exponential distribution is given as;

F(x) = P(X \leq x) = 1 - e^{-\lambda x}  , x > 0

Now, Probability that the wait time is greater than 37 minutes is given by = P(X > 37 min) = 1 - P(X \leq 37 min)

  P(X \leq 37 min) = 1 - e^{-\frac{1}{35} \times 37}        {Using CDF}

                         = 1 - 0.3474 = 0.6525

So, P(X > 37 min) = 1 - 0.6525 = 0.3474

Therefore, probability that the wait time is greater than 37 minutes is 0.3474.

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