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3 years ago

Write an inequality for the graph below

Mathematics

1 answer:

andrew11 [14]3 years ago

7 0

Answer:

0>-7

I'm not really sure last time I did this was last year

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How do you do question 6

zloy xaker [14]

Tell them what comes first and what comes last

8 0

3 years ago

To find the distance across the river, the given diagram is laid out. what is the distance rounded to the nearest meter?

kati45 [8]

Given:

Angle A = 18.6°

Angle B = 93°

Length of side AB = 646 meters

To find:

the distance across the river, distance between BC

Steps:

Since we know the measure of 2 angles of a triangle we can find the measure of the third angle.

18.6° + 93° + ∠C = 180°

111.6° + ∠C = 180°

∠C = 180° - 111.6°

∠C = 68.4°

Therefore the measure of angle C is 68.4°.

now we can use the law of Sines,

$\frac{BC}{sinA}=\frac{AB}{SinC}$

$\frac{BC}{sin(18.6)}= \frac{646}{sin(68.4)}$

$BC[sin(68.4)] = 646 [sin(18.6)]$

$BC = \frac{646*sin(18.6)}{sin(68.4)}$

$BC = \frac{646 * 0.3190}{0.9298}$

$BC = 221.63$

$BC = 222$ meters

Therefore, the distance across the river is 222 meters.

Happy to help :)

If anyone need more help, feel free to ask

7 0

3 years ago

The perimeter of the triangle below is 10x - 3<br><br> Show how to solve

bezimeni [28]

10x - 3 = x + 2 + 2x - 1
10x - 3 = 3x + 1
-3x on both sides
7x - 3 = 1
-1 on both sides
7x - 4

(Check 7x - 4 by adding them all up to equal 10x - 3)
x + 2 + 2x - 1 + 7x - 4 = 10x - 3
combine liked terms
10x - 3 = 10x - 3

Answer:
D) 7x - 4

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

Help needed ASAP will give BRAINLIEST

grin007 [14]

Answer:

Step-by-step explanation:

Recall that the sum of the (3) interior angles of a triangle <em>must</em> equate to 180.

In other words:

$(55)+(20)+(x)=180$

Where x is our unknown angle. So, we need to solve for x.

On the left, add 55 and 20:

$75+x=180$

Subtract 75 from both sides. The left side cancels:

$x=105$

Therefore, the third angle is 105 degrees.

Our answer is C.

7 0

3 years ago