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3 years ago

The volume of a spherical ball is 4500 cubic centimeters. Find the radius of the ball

Mathematics

1 answer:

babunello [35]3 years ago

8 0

Answer:

$r^{3}=4050$

r = ∛4050

Step-by-step explanation:

$V=\frac{4}{3} \pi r^{3}$

$5400=\frac{4}{3} \pi r^{3}$

$r^{3}=3(\frac{5400}{4} )$

$r^{3}=3(1350 )$

$r^{3}=4050$

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Please help if possible :)

n200080 [17]

9514 1404 393

Answer:

(a) no

(b) no

Step-by-step explanation:

The two questions are asking essentially the same thing. There is nothing that indicates DC ≅ DB or that AC ≅ AB. One (both) of these conditions is (are) required for ΔBAC to be isosceles, so that AD is both a perpendicular bisector of BC and a bisector of angle BAC.

There is not enough information to answer either question affirmatively.

8 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

What is the value of x in the equation 3/2(4x – 1) – 3x = 5/4 – (x + 2)?

Olegator [25]

i have attached a file you can see it from there.

5 0

3 years ago

Which equation shows the quadratic formula used correctly to solve 5x2 + 3x – 4 = 0 for x? x = StartFraction negative 3 plus-or-

Brrunno [24]

Answer: FIRST OPTION

Step-by-step explanation:

<h3> The missing picture is attached.</h3>

By definition, given a Quadratic equation in the form:

$ax^2+bx+c=0$

Where "a", "b" and "c" are numerical coefficients and "x" is the unknown variable, you caN use the Quadratic Formula to solve it.

The Quadratic Formula is the following:

$x=\frac{-b \±\sqrt{b^2-4ac} }{2a}$

In this case, the exercise gives you this Quadratic equation:

$5x^2 + 3x - 4 = 0$

You can identify that the numerical coefficients are:

$a=5\\\\b=3\\\\c= - 4$

Therefore, you can substitute values into the Quadratic formula shown above:

$x=\frac{-b \±\sqrt{b^2-4ac} }{2a}\\\\x=\frac{-3 \±\sqrt{(3)^2-4(5)(-4)} }{2(5)}$

You can identify that the equation that shows the Quadratic formula used correctly to solve the Quadratic equation given in the exercise for "x", is the one shown in the First option.