What angle measure A would meet the requirements so that sin A = 0.625

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:

$P(X$

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3 years ago

Is 1/3 bigger than 1/2

daser333 [38]

No, 1/2 is bigger than 1/3 because:-

1/2 = 3/6
1/3 = 2/6

Compare

3/6 is larger, which means 1/2 is larger since they both equal each other.

Answer: 1/2 is greater than 1/3

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3 years ago

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Find the area of the figure.<br> 14 m<br> 5 m<br> 16 m

Misha Larkins [42]

Answer:

Step-by-step explanation:

Seperate the shape into two. 16x2=32cm². 12x11=132cm²

So altogether it would = 164cm²

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3 years ago

Straight angles Are extremely important in geometry. When two lines intersect , they form multiple angles. In the diagram below,

rusak2 [61]

When two straight lines intersect, the vertical opposite angles intersect. the other two angles are also equal. Let the known angle be x, then the other two adjacent angles are obtained subtracting twice of x from 360 and dividing the result by 2.

Therefore, the table can by filled as follows:

Row 1:

Given <GEF = 120°

<FEM is adjacent to <GEF, thus
$\angle FEM= \frac{360-2(120)}{2} \\ \\ = \frac{360-240}{2} = \frac{120}{2} =60^o$

<MEH is vertically opposite to <GEF and thus is equal to <GEF. Thus <MEH = 120°

<HEG is vertically opposite to <FEM and thus is equal to <FEM. Thus <HEG = 60°.

Row 2:

Given <MEH = 150°

<MEH is vertically opposite to <GEF and thus is equal to <GEF. Thus <GEF = 150°

<FEM is adjacent to <GEF, thus
$\angle GEF= \frac{360-2(25)}{2} \\ \\ = \frac{360-50}{2} = \frac{310}{2} =155^o$

<HEG is vertically opposite to <FEM and thus is equal to <FEM. Thus <HEG = 30°.

Row 3:

Given that <FEM = 25°

<FEM is adjacent to <GEF, thus
$\angle GEF= \frac{360-2(25)}{2} \\ \\ = \frac{360-50}{2} = \frac{310}{2} =155^o$

<MEH is vertically opposite to <GEF and thus is equal to <GEF. Thus <GEF = 155°

<HEG is vertically opposite to <FEM and thus is equal to <FEM. Thus <HEG = 25°.

Row 4:

Given that <HEG = 45°

<HEG is adjacent to <GEF, thus
$\angle GEF= \frac{360-2(45)}{2} \\ \\ = \frac{360-90}{2} = \frac{270}{2} =135^o$

<HEG is vertically opposite to <FEM and thus is equal to <FEM. Thus <FEM = 45°

<MEH is vertically opposite to <GEF and thus is equal to <GEF. Thus <GEF = 135°.

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3 years ago

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Please help ! It's timed! (More questions)

Oksi-84 [34.3K]

Answer:

Q2 : y= -2x+9

Q4: y= 2x+8

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3 years ago