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aliina [53]
2 years ago
15

How many inches are there in 10 miles?

Mathematics
1 answer:
Arte-miy333 [17]2 years ago
8 0
There are 633600 inches in 1 mile
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Pls help thank you:)​
labwork [276]

Answer:

|2x-5|<3

Step-by-step explanation:

.............

4 0
2 years ago
An airplane takes off 12.5 miles south of a city and flies due north at a constant speed of 170 miles per hour. What is the plan
atroni [7]

Answer:

Option B

Step-by-step explanation:

Speed is distance moved per unit tine and is expressed as s=d/t

Making d the subject then

d=st

Where d is distance, s is speed and t is time taken in time.

Conversion

1 hour has 60 minutes hence 45 mins converted to hours will be 45/60=0.75

Substituting 0.75 hours for t and 170 mph for s then

d=0.75*170=127.5 miles due North

However, tge plane had initially moved 12.5 miles hence the relative distance due North will be 127.5-12.5=115 miles due North

7 0
3 years ago
Which value of x will make the following statement false?
wlad13 [49]
Im not for sure but i would say 2003-01-01-00-00-00_files
5 0
3 years ago
90 is 25 percent of what number?
Romashka [77]
25% is equal to 1/4 so that means 90 is 1 out of 4 peaces of that number so you are ganna want to multiplay 90 X 4 to get your answer
3 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
3 years ago
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