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4 years ago

How do you do all of these

Mathematics

1 answer:

aev [14]4 years ago

6 0

You just simply the answer

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PLEASE HELP IM TAKING A TEST

quester [9]

Answer:

Step-by-step explanation:

Ok bet

what do u need

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3 years ago

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Trying to get missing assignments done

Lubov Fominskaja [6]

Answer:

a-3

b-2/8 or 1/4 or 25%

Step-by-step explanation:

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3 years ago

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Find the 53rd term of the arithmetic sequence -12, -1, 10

RSB [31]

440 is the answer

Tn=a+(n-1)d

Tn=53

a= -12

n=53

d= 11

-12+(53-1)11 = 440

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3 years ago

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A mermaid has a terrarium that measures 42 inches × 10 inches × 9 inches. She also has a box

Serjik [45]

Answer:

105 minutes

Step-by-step explanation:

The volume of the terrarium is ...

V = LWH

V = (42 in)(10 in)(9 in) = 3780 in³

The volume of a box is ...

V = (3 in)(3 in)(3 in) = 27 in³

Then the number of boxes it takes to fill the terrarium is ...

(3780 in³)/(27 in³/box) = 140 boxes

The mermaid wants to fill it 3/4 full, so only needs 3/4 this number of boxes. That is 105 boxes. At the rate of 1 per minute, ...

it will take 105 minutes to fill 3/4 of the terrarium

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3 years ago

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

marshall27 [118]

Answer:

1. 0.0000454

2. 0.01034

3. 0.0821

4. 0.918

Step-by-step explanation:

Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,

$X \sim Poi(\lambda = 10)$

1. Requires us to compute

$P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454$

2. We need to compute $P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3)$

$P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454$

$P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227$

$P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757$

$P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034$

3. The expected no. of arrivals in a 15 second period is = $10 \times \frac{1}{4} = 2.5$ . So if Y be the random variable denoting number of passengers arriving in 15 seconds,

$Y \sim Poi(2.5)$

$P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821$

4. Here we use the fact that Y can take values $0,1, \dotsc$ . So, the event that "Y is either 0 or $\geq$ 1" is a sure event ( i.e it has probability 1 ).

$P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918$

3 0

3 years ago