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andrew-mc [135]
3 years ago
15

What is 1.2 simplified

Mathematics
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

it is 1 2/10 but simplified to 1 1/5

Step-by-step explanation:

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The random variable X is exponentially distributed, where X represents the time it takes for a person to choose a birthday gift.
NeX [460]

Answer:

0.674

Step-by-step explanation:

If the random variable X is exponentially distributed and X has an average value of 25 minutes, then its probability density function (PDF) is

\bf f(x)=\frac{1}{25}e^{-x/25}\;(x\geq 0)

and its cumulative distribution function (CDF) is

\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/25}

So, <em>the probability that X is less than 28 minutes is </em>

\bf P(X\leq 28)=1-e^{-28/25}=1-e^{-1.12}=0.674

8 0
3 years ago
Compare the fraction 3/5 to 1/2 and then compare 2/6 to 1/2 .use symbols in your answer.
Feliz [49]
3/5 > 1/2 ~ because 3/5 is 60% of 1 and 1/2 is 50% of 1 
2/6<1/2 ~ because 2/6 is 33% of 1 and 1/2 is 50% of 1

Hope I helped!
~ Zoe
3 0
3 years ago
How many solutions are there to the following system of equations?. . 2x – y = 2. –x + 5y = 3. A. 1 B. 0 C. 2 D. infinitely many
krek1111 [17]
  2 x - y = 2
- x + 5 y = 3   / *2
----------------------
  2 x  -  y  =  2
+
- 2 x + 10 y = 6
-----------------------
           9 y = 8,  y = 8/9
2 x - 8/9 = 2
2 x = 2 + 8 /9
2 x = 26/9
x = 13/9
( x, y ) = ( 13/9,  8/9 )
Answer: A ) 1 solution. 
4 0
3 years ago
Read 2 more answers
In the year 2000, the population of a city was 600,000 citizens. The population increases at a rate of 1.8% per year.
Alona [7]
Y=600000(1+1.8%)^(x-2000)
So the population in 2012
Y=600,000×(1+0.018)^(2,012−2,000)
Y=743,232
6 0
3 years ago
Read 2 more answers
How could you find the y-coordinate of the midpoint of a vertical line segment with endpoints at (0,0) and (0,-12)
Oduvanchick [21]
I think it should be -6, because line segment equals 12 and it's midpoint will be 12/2 = 6, so you must go down by 6 from (0,0) point on y axis which is 0-6=-6

So the point will be (0, -6) 
8 0
3 years ago
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