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Lina20 [59]
3 years ago
14

What is the slope of the line?

Mathematics
1 answer:
natima [27]3 years ago
5 0

Answer:

(2,1)

Step-by-step explanation:

Because it rises by 2 and runs by 1

You might be interested in
Lillian had 2 two-liter bottles of soda, which she distributed equally between 10 glasses.
alexgriva [62]

<u><em>Answers:</em></u>

a. \frac{2}{5} liters/glass

b. 0.4 liters/glass

c. 400 ml/glass

<u><em>Explanation:</em></u>

<u>Part a:</u>

We are given that two bottles each holding 2 liters of soda

This means that the total amount of soda = 2 * 2 = 4 liters

To get the amount of soda in each glass, we will simply divide the amount of soda by the number of glasses (given the number of glasses is 10)

<u>Therefore:</u>

Amount of soda in each glass = \frac{4}{10} = \frac{2}{5} liters/glass

<u>Part b:</u>

We are now asked to express the amount in fraction. The simplest way to do this is to make the denominator a multiple of 10. The number of zeroes in the denominator will be equal to the number of digits after the decimal point

<u>Therefore:</u>

\frac{2}{5}=\frac{4}{10}=0.4 liters/glass

<u>Part c:</u>

Now, we know that 1 liter is equivalent to 1000 ml

<u>Therefore:</u>

0.4 liters = 0.4 × 1000 = 400 ml/glass

Hope this helps :)

6 0
3 years ago
Read 2 more answers
*ASAP*
In-s [12.5K]
If i can see it right, it would be A because it starts at 30, then rises as the number of rooms increases.
4 0
3 years ago
An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me
kirill115 [55]

Answer:

a) \lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

b) P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

c)  e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

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 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"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given \mu = 0.5 and we can solve for the parameter \lambda like this:

\frac{1}{\lambda} = 0.5

\lambda =2

So then X(t) \sim Poi (\lambda t)

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

\lambda_1 = 2*2 = 4

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

X(2) \sim Poi (4)

And the expected value is E(X) = \lambda =4

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

P(X(2) >6)

And we can use the complement rule like this

P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]

And we can solve this like this using the masss function:

P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]

And we got: P(X(2) >6) =1-0.889=0.111

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

T \sim Exp(\lambda=2)

The probability of no arrival during a period of duration t is given by:

f(T) = e^{-\lambda t}

And we want to find a value of t who satisfy this:

e^{-2t} \leq 2

We can apply natural log in both sides and we got:

-2t \leq ln(0.2)

If we multiply by -1 both sides of the inequality we have:

2t \geq -ln(0.2)

And if we divide both sides by 2 we got:

t \geq \frac{-ln(0.2)}{2}

t \geq 0.8047

And then we can conclude that the time period with any load would be 0.8047 years.

3 0
3 years ago
The polygons in each pair are similar. Find the scale factor of the smaller figure to the larger figure.
Step2247 [10]
7 literally just 7......
7 0
2 years ago
2+2?<br> I need help this is really hard helpppp
Andrews [41]
The answer is 4 . 2 + 2 = 4 . hope i helped uuuu lol
4 0
3 years ago
Read 2 more answers
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