29/4 as a percentage

2% of the students at Hamilton Middle School have red hair. There are 700700 students at Hamilton Middle School. How many studen

dsp73

700,700 times 0.02 = 14,014 students

4 0

3 years ago

What is 60% of $49000?

katen-ka-za [31]

Suppose you have a Kohls coupon of $49000 and you want to know how much you will save for an item if the discount is 60 percent.

Solution:

Replacing the given values in formula (a) we have:

Amount Saved = Original Price x Discount in Percent / 100. So,

Amount Saved = 49000 x 60 / 100

Amount Saved = 2940000 / 100

Amount Saved = $29400 (answer).

In other words, a 60% discount for a item with original price of $49000 is equal to $29400 (Amount Saved).

Note that to find the amount saved, just multiply it by the percentage and divide by 100.

4 0

3 years ago

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It takes 32 hours for a motorboat moving downriver to get from pier A to pier B. The return journey takes 48 hours. How long doe

lina2011 [118]

Answer:

Time taken by the un powered raft to cover this distance is T = 192.12 hr

Step-by-step explanation:

Let speed of boat = u $\frac{km}{hr}$

Speed of current = v $\frac{km}{hr}$

Let distance between A & B = 100 km

Time taken in downstream = 32 hours

$32 = \frac{100}{u +v}$

$u + v = \frac{100}{32}$

u + v = 3.125 ------ (1)

Time taken in upstream = 48 hours

$48 = \frac{100}{u -v}$

$u -v = \frac{100}{48}$

u - v = 2.084 ------- (2)

By solving equation (1) & (2)

u = 2.6045 $\frac{km}{hr}$

v = 0.5205 $\frac{km}{hr}$

Now the time taken by the un powered raft to cover this distance

$T = \frac{100}{v}$

Because un powered raft travel with the speed of the current.

$T = \frac{100}{0.5205}$

T = 192.12 hr

Therefore the time taken by the un powered raft to cover this distance is

T = 192.12 hr

3 0

3 years ago

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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

I need help with this and the first person who can answer this correctly gets a BRANLIST

Vesna [10]

Answer:

your answer is

A

6 0

3 years ago

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