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3 years ago

Answer C,D,E,FSnsjma

Mathematics

1 answer:

Ulleksa [173]3 years ago

6 0

Answer:

f) $97° = m∠6$

e) $83° = m∠5$

d) $83° = m∠4$

c) $97° = m∠3$

b) $97° = m∠2$

a) $83° = m∠1$

Step-by-step explanation:

m∠5 ≅ m∠4; Angle Correspondence Theorem
m∠5 ≅ m∠1; Alternate Interior Angles Theorem
m∠6 ≅ m∠2; Alternate Exterior Angles Theorem
180° = m∠2 + m∠1; Definition of Linear Pair [Supplementary Angles]
180° = m∠5 + m∠3; Definition of Linear Pair [Supplementary Angles]

I am joyous to assist you anytime.

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

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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:

$P(X$

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3 years ago

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natita [175]

To find c, you must isolate it.
To do this, you must divide both sides by 5/7, since that is being multiplied by c and you must do the inverse to it to cancel it out in order to leave c by itself.

5/7c ÷ 5/7 = c

13/14 ÷ 5/7
To divide fractions, follow these steps:
Step 1- Turn the second fraction, 5/7 in this case, into its reciprocal. This means swapping the places of the numerator and denominator.
5/7 reciprocal = 7/5

Step 2- multiply the original first fraction and reciprocal second fraction.
13/14 • 7/5
13 • 7 = 91
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13/14 ÷ 5/7 = 91/70

Step 3- Simplify if possible.
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Since 70 can go into 90, you can turn this into a mixed number.
1 and 21/70
Now simplify 21/70.
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21 ÷ 7 = 3
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So the answer is c = 1.3, or 1 and 3/10.

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