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makvit [3.9K]
2 years ago
13

Martin will earn $20 per hour at a new job. During training he will earn $15 per hour. What percent of Martin’s regular hourly r

ate well he earned during training?
Mathematics
1 answer:
Contact [7]2 years ago
4 0

Answer:

75%

Step-by-step explanation:

To find the percentage, you would divide the training hourly rate by Martin's regular hourly rate and multiply the quotient by 100

If martin normally earns $20 and he earns $15 during training, that means you would divide 15 by 20.

15/20=.75 and .75x100=75, so Martin earns 75% of his regular hourly rate during training

Please mark as brainliest ;)

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Given Arc KM measures 130 degrees, what is the measure of angle MLK?
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Pete can buy 4 cokes for $5 or 10 cokes for $12.50. How much will he would pay for 7 cokes?
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The time T required to repair a machine is an exponentially distributed random variable with mean 10 hours.
Firdavs [7]

It can be expected about 36.79% of chance that repair time exceeds

The probability that a repair time exceeds  15 hours is 0.3679

What is the exponential distribution?

It explains about the time between events or the distance between two random events is termed the exponential distribution. Here, the occurrence of the events is continuous and also independent. Moreover, the average rate is constant.

The cumulative distribution function of T is obtained below:

From the information given, let the random variable T be the required time to repair a machine follows exponential distribution with parameter λ
with mean. 1/2 hours

That is,  E(x) =  1/2 hours.
The parameter of the random variable T is,
E(x) =  1/λ
λ = 1/E(x)
= 1/(1/2)
= 2

The probability density function of T is,
f(t) = \left \{ {{2e^{-2t} \ \ \ t > 0}  \  \atop {0} \ \ \ elsewhere} \ \right.
The cumulative density function of T is,
FT(t) = P(T <= t)

= 1 - e^{- \lambda t}

= 1 - e^{- 2t}
The CDF of T is,
P(T <= t)  = 1 - e^{- 2t}    0 <= T <= ∞
        = 0          otherwise
to obtain the probability that a repair time exceeds

1/2 hours.
(a) The probability that a repair time exceeds 1/2 hours.
From the given information, the CDF of T is,
P(T <= t)  = 1 - e^{- 2t}    0 <= T <= ∞
        = 0          otherwise

The required probability is,
P(T <= 1/2)  = 1 - P(T <= 1/2)
        = 1 - [  = 1 - e^{- 2(1/2)} ]
       = e^{-1}
= 0.3679

om total probability. It can be expected about 36.79% of chance that repair time exceeds

P(X => x)  = 1 - P(X < x)
to obtain the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours.

(b), The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained below:
From the given information, the CDF of T is,
P(T <= t)  = 1 - e^{- 2t}    0 <= T <= ∞
        = 0          otherwise
The required probability is,
P = P(T => 12.5∩T>12) / P(T>12)
= e^{- 25 + 24}

= e^{- 1}
= 0.3679
The probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours is obtained by dividing the
P = P(T => 12.5∩T>12) / P(T>12)
with
P(T>12).

It can be expected about 36.79% of chance that a repair takes at least 12.5 hours given that its duration exceeds 12 hours.

Hence, It can be expected about 36.79% of chance that repair time exceeds,

The probability that a repair time exceeds  15 hours is 0.3679

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brainly.com/question/22692312
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