Ask question

Ask question

All categories

3 years ago

14

You score 84 on the first test, 95 on the second test, and 87 on the third test.

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

91

Step-by-step explanation:

You might be interested in

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

Rewrite the quadratic funtion from standard form to vertex form. <br><br> F(x)=2x^2-20x+26

Olegator [25]

F(x)=2(x-5)^2-24 simple algebra

3 0

2 years ago

sashaice [31]

Answer: 187.7 mph

Step-by-step explanation:

v = 125 mph

t = 1.2 min

v2 = ?

t2 = 0.8 min

We can use the rule of 3 here. It's the inverse rule of 3 because as time decreases, speed increases.

$\frac{125mph}{x} =\frac{1.2min}{0.8min}$

$x = \frac{(125)(1.2)}{0.8}$

$x=187.5mph$

7 0

3 years ago

What is 7 times 9 in multiplti

NemiM [27]

The answer is 63. 9 times 7 is 63

7 0

3 years ago

Read 2 more answers

I'm thinking of two numbers. The first plus twice the second equals 11. Twice the first plus three times the second equals 50. W

valentinak56 [21]

2x + 4y - 2x - 3y = 22 - 50

y = -28

substitute into the 1st equation to find x

3 0

3 years ago