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anastassius [24]

3 years ago

10

Jack wants to purchase a Wallet for $33. He currently has $18 . By how much money will he be in debt if he decides to borrow the

difference ?

1 answer:

Semmy [17]3 years ago

4 0

Answer:

$15 in debt

Step-by-step explanation:

Subtract 33 from 18

18 - 33

= -15

So, he will be $15 in debt

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5. Find the value of x.

goldfiish [28.3K]

Answer:

x = 70

Step-by-step explanation:

Interior angles of a 4 sided shape = 360

80+70 = 150

360-150 = 210

210/3 = 70

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3 years ago

What is the range of the function represented by the table of values below?

Viefleur [7K]

The range is the interval of y values.
Smallest y values is -9 and the highest y values is 9.
So the range is $-9 \leq y \leq 9$

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3 years ago

Every five years in March, the Population of a certain town is recorded. In 1995, the town had a population of 4,500 people. Fro

alexgriva [62]

4,500 plus 15% is 5,175 minus 4% is 4,968 so your answer is 4,968.

5 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

Write the equation of a line that includes the point (-4 , 0) and has a slope of -3/5 in slope intercept form

erastovalidia [21]

Answer: " y = (-3/5)x − 4 " .
______________________________________________

4 0

3 years ago

Read 2 more answers