How do I solve this

Freya, a post office clerk, uses 2 yards, 1 foot, 7 inches of twine to wrap each damaged carton she repairs. She has to repair 1

sattari [20]

Per damaged carton the clerk uses 7 feet 7 inches of twine.
or
91 inched total.
multiply 91 times the amount of cartons needed (17) your answer is 1547

5 0

3 years ago

Read 2 more answers

I need this fast thank you! :)

Alex Ar [27]

Answer:

C

Step-by-step explanation:

Hope this helps! Good Luck!

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

What is the probability that a randomly thrown dart that lands within the rectangle lands within a shaded region? All of the cir

marusya05 [52]

D is the answer because the circle has no sides so it’s 157/200

8 0

2 years ago

Help please will give B

Vinvika [58]

Hi there! Hopefully this helps!

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<em>Remember: When the numerator and denominator both are positive integers or both are negative integers, it is a positive rational number. When either the numerator or the denominator is a negative integer, it is a negative rational number.</em>

(-1/-3) Positive Rational Number.

(+1/-3) Negative Rational Number.

(-2/+3) Negative Rational Number.

(+1/+3) Positive Rational Number.

(+2/-3) Negative Rational Number.

(-1/+3) Negative Rational Number.

(-2/-3) Positive Rational Number.

(+2/+3) Positive Rational Number.

4 0

3 years ago