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kondor19780726 [428]
3 years ago
13

What is 7,060,268,214 rounded to the nearest hundred million

Mathematics
2 answers:
aniked [119]3 years ago
5 0
7,100,000,000 hope i helped you out in some type of way :) 
tatiyna3 years ago
3 0
7,060,268,214 rounded to the nearest hundred million is 7,100,000,000
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Firlakuza [10]

Ya I believe that is the answer. I got the exact same thing.

6 0
3 years ago
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Find f(5) please explain
kirza4 [7]

Answer:

-1

Step-by-step explanation:

A function is represented by the following symbol f(x) where f(x) further represents y like this ,

y=f(x)

so in this question they are asking what is f(5) which means when the value of x is 5 what is the value of y?

We can see in the graph when x = 5 the value of y is -1 to be exact so

f(5) = -1

or

y=f(x)\\-1 = f(5)\\

8 0
3 years ago
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
3 years ago
A polynomial P(x) and a divisor d(x) are given. Use long division to find a quotient Q(x) and the remainder R(x). Express P(x) i
Vilka [71]
It's a bit hard to explain but I hope this helps

5 0
3 years ago
How do I find ac in this problem
suter [353]

7 / x+4 = 5/x+1 =


5x+20 = 7x+7 =

20=2x+7

13=2x

x=13/2 = 6.5

6.5+4 = 10.5

6.5+1 =7.5

10.5+7.5 = 18

length of AC = 18

8 0
3 years ago
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