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4 years ago

What is the equivalent ration for 12/30

Mathematics

2 answers:

Novosadov [1.4K]4 years ago

7 0

(12/30) / 6 = 2/5
You cannot reduce 5ths anymore therefore it’s the smallest equivalent

TiliK225 [7]4 years ago

5 0

Answer:

24/60 or 6/15

Step-by-step explanation:

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Please Answer ASAP! Answer MUST include an explanation in order to receive points and the Brainliest answer. Thank you.

miss Akunina [59]

Probablity is (desired outcomes)/(total possible outcomes)

total possible outcomes is 50

find the desired oucomes
les than 10
those ar
1,2,3,4,5,6,7,8,9
9 numbers less than 10

OR
multipule of 12
12,24,36,48
4 of them

9+4=13
13=total possible

13/50
C

7 0

4 years ago

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Brrunno [24]

Answer:

Thank you

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

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

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

4 years ago

The cost of shoes is 200 the code was raised by 5% what is the price now

adoni [48]

5% of 200 is 10

200 + 10 is 210

4 0

4 years ago

What is the midpoint of a segment with These endpoints (-6,-9) and (16,5)

lora16 [44]

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{-9})\qquad (\stackrel{x_2}{16}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 16 -6}{2}~~~ ,~~~ \cfrac{ 5 -9}{2} \right)\implies \left( \cfrac{10}{2}~~,~~\cfrac{-4}{2} \right)\implies (5~~,~-2)$

7 0

3 years ago