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irga5000 [103]
3 years ago
7

Sound like all the situations that I’ll take three hours

Mathematics
1 answer:
viktelen [127]3 years ago
5 0
Time = Distance / Speed

1. 240/60 = 4 Hours
2. 6/3 = 2 Hours
3. 90/30 = 3 Hours

Therefore the correct answer is 3.
You might be interested in
The executives in a company are four women and three men. Three are to be chosen at
Lisa [10]
The correct answer is A i hope this helps
4 0
3 years ago
Which of the following graphs could be the graph of the function f(x)=-0.08x(x^2-11x+18)?
MariettaO [177]

Answer:

option C

Step-by-step explanation:

f(x)=-0.08x(x^2-11x+18)

Factor the parenthesis and find out the x intercepts

(x^2-11x+18)

We find out two factors whose product is 18 and sum is -11

-9  and -2 gives us product 18

-9  -2 gives us sum -11

(x^2-11x+18) =(x-9)(x-2)

f(x)=-0.08x(x^2-11x+18)

f(x)= -0.08x(x-9)(x-2)

set f(x) =0  and solve for x

-0.08x(x-9)(x-2) =0

-0.08x =0, so x=0

x-9 = 0 , so x=9

x-2 =0 , so x=2

x intercepts are (0,0) (2,0)  and (9,0)

x intercepts are not repeating

so the graph crosses x -axis at x=0, 2,9

option C is correct

The graph is attached below4


5 0
3 years ago
Read 2 more answers
What is the area of this tile?
8_murik_8 [283]

Answer:

The area is 480 inches

10 × 8 = 80

80 × 6 = 480

8 0
4 years ago
Help me solve this please
Semenov [28]

\huge\underline{ \underline{ \bf{Answer࿐} }}

parallel sides measure :

  • 10 ft -(a)
  • 12 ft. -(b)

Area :

  • 44 ft²

height (h) = x

___________________

\boxed{Area =  \frac{1}{2}  \times (a + b) \times h}

  • \dfrac{1}{2}  \times (10 + 12) \times x = 44

  • \dfrac{1}{2}  \times 22 \times x = 44

  • 11x = 44

  • \boxed{ 4 \:  \: ft}

Therefore, height of Trapezoid is 4 ft

_____________________________

\mathrm{ \#TeeNForeveR}

5 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
4 years ago
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