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kati45 [8]
4 years ago
9

Which fraction is equivalent to 0.65? A) 5/13 B) 13/20 C) 19/25 D) 27/35

Mathematics
2 answers:
Irina-Kira [14]4 years ago
8 0
The answer is B. 13/20

To do this, you may need a caculator if you want to. Then put 13 divided by 20 as a fraction and you’ll get 0.65
Ulleksa [173]4 years ago
4 0
Rewrite the decimal number as a fraction with 1 in the denominator
0.65=0.651
0.65
=
0.65
1
Multiply to remove 2 decimal places. Here, you multiply top and bottom by 102 = 100
0.651×100100=65100
0.65
1
×
100
100
=
65
100
Find the Greatest Common Factor (GCF) of 65 and 100, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 5,
65÷5100÷5=13/20
65
÷
5
100
÷
5
=
13/20
Therefore
=13/20
X
=
13/20
In conclusion,
0.65=13/20

The answer is B
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Choose the correct equivalent expression that includes a set of parentheses so that the value of the expression is 6.
kicyunya [14]

Answer:20- 2 x (5 +2)

Step-by-step explanation:

20 - 2 x (5 + 2)

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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

To learn more about the product of the fraction visit,
brainly.com/question/22692312
#SPJ4

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s2008m [1.1K]
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