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Vikentia [17]
3 years ago
5

When will the distance from A to B be negative?

Mathematics
1 answer:
Aneli [31]3 years ago
6 0
The distance from A to B will be negative when the line is trending downward from left to right.
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How to find the value of a+b+c given that a=4,b=3 and c= -2
Alexus [3.1K]

Answer: 5

Step-by-step explanation: 4 + 3 + -2 = 5

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3 years ago
The area of this parallelogram is 120 ft. Find the value of h
julsineya [31]

are you going to show a picture??? its impossible to answer otherwize

4 0
3 years ago
Find the midpoint of the line segment JK if j (4,6) and k(0,-4)
Snowcat [4.5K]

Answer:

(2,1)

General Formulas and Concepts:

Order of Operations: BPEMDAS

Midpoint Formula: (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Step-by-step explanation:

<u>Step 1: Define points</u>

J (4, 6)

K (0, -4)

<u>Step 2: Find midpoint</u>

  1. Substitute:                         (\frac{4+0}{2},\frac{6-4}{2})
  2. Add/Subtract:                    (\frac{4}{2},\frac{2}{2})
  3. Divide:                               (2,1)
4 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
8x+4(x-3)=4(6x+4)-4 please help
kvasek [131]

Answer:

x = -2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

8x + 4(x - 3) = 4(6x + ) -4

8x +(4)(x) + (4)(-3) = (4)(6x) + (4)(4) + -4 (distribute)

8x + 4x + -12 = 24x + 16 + -4

(8x) + (4x) +(-12) = (24x) + (16 + -4) (combine like terms)

12x + -12 = 24x + 12

12x -12 = 24x + 12

Step 2: Subtract 24x from both sides.

12x -12 -24x = 24x + 12 -24x

-12x -12 = 12

Step 3: Add 12 to both sides.

-12x - 12 + 12 = 12 + 12

-12x = 24

Step 4: Divide both sides by -12.

-12x/-12 = 24/-12   <u>x = -2</u>  

HOPE THIS HELPED!!!

5 0
3 years ago
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