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3 years ago

How to reduce 10/4 i don’t know how to

2 answers:

5 0

10 and 4 are both divisible by 2. so divide 10 by two and you would get 5 but what you do to the top you do to the bottom. so divide 4 by 2 and you get 5/2 or 2 1/2

vredina [299]3 years ago

3 0

answer: 10/4 reduced is 2.5/1 or 2.5.

work: what we are doing is trying to divide ten into groups of four. how many times does 10 fit into four? well, look at the multiples of four, 4, 8, 12... we can see that 8 fits into ten, but 12 doesn't. that means two whole fours can fit into ten (since 4 x 2 = 8) so we still have a 2 left over from the ten. since 2 is one half of four, we can fit in another .5. so, ten can be fit into 2.5 groups. hope this helps, have a great day!

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What’s 93/108 in simplest form

mafiozo [28]

31/36 is the correct answer

4 0

3 years ago

Read 2 more answers

Round or compatible 198 add 727 to the estimate sum

uranmaximum [27]

Lets break this down here.

First- 198 rounded would be 200
Then- we add 727 which would be 927
if you were to round that it would be 930

3 0

3 years ago

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$