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3 years ago

FIRST CORRECT ANSWER GET BRAINLIEST

Mathematics

2 answers:

erma4kov [3.2K]3 years ago

7 0

Omg it is football!!!!!!!!

Yakvenalex [24]3 years ago

3 0

Plz give TheStudyGod Brainliest!

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A skydiver in a computer game free-falls from a height of 3000 meters at a

mylen [45]

H= -55t + 3000. Hope this helps!

8 0

2 years ago

What is the value of the expression 3a+b2 when a=14 and b=32 ?

soldi70 [24.7K]

Answer:

106

Explanation:

3a+b2

Substitute the values for the variables:

3(14) + 2(32)

Solve:

3(14) + 2(32)

42 + 64

106

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

4 years ago

A charity fair raised $6,000 by selling 500 lottery tickets. There were two types of lottery tickets: A tickets cost $10 each, a

lara [203]

Answer:

it would be E

Step-by-step explanation:

20x60= 4800

480x10= 1200

4800+1200= 6000

6 0

3 years ago

Plz answer all corectly will mark brainlest plz answer within 1 hour

denis23 [38]

Answer:

for the first one I got y=-2x-2 the second part i got y=-x-1 or -1x-y-1=0 Correct me if I am wrong

Step-by-step explanation:

Use the slope formula

-4-4 =-8

/ =-2

1 - -3=4

2. The second part use this formula y-y1=m(x-x1)

y--4=-1(x-3)

y+4=-x+3

-4 -4

y+0=-x-1

y=-x-1 or -1x-y-1=0 I am not sure

6 0

3 years ago

****FIRST CORRECT ANSWER GET BRAINLIEST****

FIRST CORRECT ANSWER GET BRAINLIEST