I need help with math can someone help me I’ll show the question?

A tennis ball has a diameter of 2.7 inches. What is the approximate volume of the tennis ball

jonny [76]

Answer:

volume = 10.2 in³

Step-by-step explanation:

r = 2.7/2 = 1.35

volume = 4/3πr³ = 4/3(3.14)(1.35³) = 10.2 in³

4 0

1 year ago

Kenneth loaned $1,200 to his niece to buy a computer. Two years later, she paid him back the $1,200 along with an interest of $9

insens350 [35]

Answer:

The annual interest rate was 4%

Step-by-step explanation:

1. Let's review the data given to us for solving the question:

Loan to Kenneth's niece to buy a computer = US$ 1,200

Duration of the loan = 2 years

Niece paid back to Kenneth = US$ 1,296

Annual interest rate = x%

2. Let's find the annual interest rate after 2 years, using the following formula:

FV = PV * (1 + x)

PV = Loan to Kenneth's niece to buy a computer = US$ 1,200

FV = Niece paid back to Kenneth = US$ 1,296

rate (r) = x

Replacing with the real values:

1,296 = 1,200 * (1 + x)

1.296/1.200 = (1 + x) (Dividing by 1,200 at both sides)

1.08 = 1 + x

1.08 - 1 = x (Subtracting 1 at both sides)

0.08 = x

This is the interest rate for two years, for finding the annual interest rate we divide it by 2, this way:

0.08 / 2 = 0.04 * 100 = 4%

7 0

2 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

2 years ago

12 picks in all . 1 out of 4 picks are orange . the rest are green . How many picks are orange ?

makvit [3.9K]

For every 4 picks there is 1 orange pick, so 12/4 should be fine

7 0

3 years ago

Events A and B are mutually exclusive. Also P(A) =0.2 and P(B) = 0.37. What is P(A or B)

Alchen [17]

Answer:

0.074

Step-by-step explanation:

cause in probability when two events are exclusive we use multiplication rule

7 0

2 years ago