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3 years ago

What is .064 as a fraction?

Mathematics

2 answers:

Over [174]3 years ago

8 0

Answer:

8/125

Step-by-step explanation:

Eight over one hundred twenty five

lidiya [134]3 years ago

3 0

.064 as a fraction would be 8/125

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PLEASE HELP!!! 15 POINTS!!

tankabanditka [31]

Answer:

23.4 cm

Step-by-step explanation:

If the base of the triangle is half of that of a given side, then we need to divide the given side by 2 to get the base.

46.8

---------2

23.4

There is your answer

Hope it helped

Bye

7 0

3 years ago

Read 2 more answers

I need help on both of these please///:(

SCORPION-xisa [38]

14. The distance between the two points is 14.866.

Distance can be calculated with the following formula:

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(12-2)²+(5-(-6))²

d=√10²+11²

d=√100+121

d=√221

d=14.866

15. The distance between the two points is 20.248.

Use the same formula to find the distance.

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(4-(-3))²+(12-(-7))²

d=√7²+19²

d=√49+361

d=√410

d=20.248

4 0

4 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

Read 2 more answers

What is 1.147 , 1.15 , 1.098 , 0.884 from greatest to least

SOVA2 [1]

1.15,1.147,1.098,.884

5 0

3 years ago

Read 2 more answers

Lindsey earn 70$ for working 5 hours. How much does she earn for working 12 lawns ?

Mice21 [21]

Assuming Lindsey takes 5 hours to mow a lawn, she earns from twelve lawns
12*70=$840.

If the above assumption is not correct, @funnyork2005 needs to clarify the question.

6 0

4 years ago