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2 years ago

10

Calculate the density of a piece of wood with a mass of 80g and a volume of 90cm³

1 answer:

likoan [24]2 years ago

4 0

Answer:

0.89 g/cm³

Step-by-step explanation:

used the formula d=M/V

M is mass which is = 80g

V is volume which is = 90cm³

d = 80/90

d = 0.89 g/cm³

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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

4 years ago

Select ALL 4 ways of displaying a relation.

a_sh-v [17]

Answer:

Scatter plots

Pie graph

graph

in words

7 0

3 years ago

If x = 10, which lines are parallel? Check all that apply.

OlgaM077 [116]

Answer:

The answer would be A. 1 1/5

Step-by-step explanation:

1 8/40 Simplify to 1 1/5

48/40 = 1 8/40

The answer to this question is above.

Hope this helps.

3 0

3 years ago

Which of the following relations is a function? A. (1, 4), (-4, 2), (8, 1), (-8, 2) B. (1, 4), (-4, 6), (1, 3), (-8, 2) C. (1, 0

Len [333]

Answer:

A. (1, 4), (-4, 2), (8, 1), (-8, 2)

Step-by-step explanation:

Each x goes to only 1 y to be a function

A. (1, 4), (-4, 2), (8, 1), (-8, 2)

function

B. (1, 4), (-4, 6), (1, 3), (-8, 2)

1 goes to 4 and 3 so not a function

C. (1, 0), (-4, 3), (8, 1), (-4, 5)

-4 goes to 3 and 5 so not a function

D. (8, 1), (-4, 4), (1, 1), (8, 2)

8 goes to 1 and 2 so not a function

6 0

4 years ago

Read 2 more answers

H=16t squared +80t+96

Agata [3.3K]

First you combine the similar variables such as, 16T and 80T then you should come up with H=96t squared + 96 and then simplify!

7 0

3 years ago