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3 years ago

Anyone know how to solve a functions equation such as x^2-x-x <0

Mathematics

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

Step-by-step explanation:

$x^{2} -x-x$ 0

$x^{2} -2x$ < 0

x^2-2x+1<1

(x-1）^2<1

-1<x-1<1

0<x<2

$(x-1)^{2}$ $(x-1)^{2}$

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Find the next term if the sequence: 5, 5, 10, 30, 120, 600

adell [148]

Answer:

3600

Step-by-step explanation:

5 * 5 = 1

5 * 2 = 10

600 * 6 = 3600

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2 years ago

A display of cans in a grocery store consists of 15 rows. There are 45 cans on the bottom , 42 cans on the next row, and so on,

Alchen [17]

Answer:

n = 9

Step-by-step explanation:

20 + 18 + ... + 6 + 4

---------------------------

a = 20 ; d = -2 ; n = ?

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4 = 20 + (n-1)(-2)

-16 = (n-1)(-2)

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n-1 = 8

n = 9

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S(9) = (9/2)(20+4)

S(9) = 4.5*24

S(9) = 108

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3 years ago

Please help me! simplify. 5-[-(-20] a. -3 b.3 c.-7 d.7

frosja888 [35]

Im sorry i dont know

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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

Erin has scanned a 4 × 5-inch postcard, and she wants to reduce it to make a graphic for her Web site. She will use a scale fact

Rasek [7]

Just multiply both dimensions by 2 to make it into centimeters.

Therefore,

4*2 x 5*2 (in inches) will be:

8cm x 10cm (in centimeters)

We conclude that the dimensions of her postcard into a graphic on her web page will be 8x10 in centimeter units.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

6 0

2 years ago