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3 years ago

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Reasonableness. there are 53 students on a field trip. one chaperone is needed for every 6 students. how many chaperones are nee

ded?
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2 answers:

lesya692 [45]3 years ago

8 0

K so u would divide 53 and 6 and u would get 8 r 5 so that means 5 students can't go to the field trip so u need one more chaperone for the 5 students so u would need 9 chaperone

xxMikexx [17]3 years ago

7 0

To find out how many chaperones are needed we need to know the multiples of 6.
6<span>,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96,102
Counting how many times 6's go into 53, we get 9.
We will need 9 chaperones!
I hope that helps!</span>

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The five members of the Traynor family each buy train tickets. During the train ride, each family member buys a boxed lunch for

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Let's assume

the price of each train ticket is x

we are given

total number of family members =5

so, total cost of ticket = total number of family members*the price of each ticket

so, total cost of ticket =5*x

so, total cost of ticket =5x

now, we have

each family member buys a boxed lunch for $6.50

so, total cost of lunch box = total number of family members*price of each lunch box

so, total cost of lunch box = 5*6.50

total cost of trip = total cost of ticket+total cost of lunch box

we are given

total cost of trip is $248.50

now, we can plug values

and we get

$248.50=5x+5*6.50$

now, we can solve for x

$248.50=5x+32.5$

$248.50-32.5=5x+32.5-32.5$

$216=5x$

$\frac{5x}{5} =\frac{216}{5}$

$x=43.2$

so,

the price of each train ticket is $43.2..........Answer

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Step-by-step explanation:

First, we realise

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Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

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c) $P(X \geq 8) = 1-P(X$

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Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

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a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

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$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

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$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

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c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

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The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

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