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3 years ago

6

4 friends evenly divided up an n-slice pizza. One of the friends, Harris, ate 1 fewer slice than he received. How many slices of

pizza did Harris eat? Write your answer as an expression. Pls answer fast I'm taking a test!!

1 answer:

atroni [7]3 years ago

6 0

Answer:

7

Step-by-step explanation:

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Jason helped his dad put plastic and aluminum items into the recycling container. The container is a large cube with a length, w

igor_vitrenko [27]

Answer:

its 216 cubic feet.

Step-by-step explanation:

6x6x6 is 216

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3 years ago

Identify the terms themselves for the following expression. (separate ur terms with a comma)

Zinaida [17]

3a, 5b, -6y, 10 hope this helps

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3 years ago

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Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

Please help me! I will make you the brainliest! Don't answer if you don't know

Helen [10]

Table C is a function, because none of the numbers in the X column repeat. (All the other ones do.):-)Hope I could help!!

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3 years ago

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PLEASE HELP!! ASAP!!

mr Goodwill [35]

Answer:

D and E

Step-by-step explanation:

A vertical asymptote in this case would mean that the denominator is equal to zero, which is undefined. You can factor the denominator to get you the fraction:

x+2 / (x+6)(x-4) = 0

Then you know that to equal zero, x would have to equal either -6 or 4, which is E and D.

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3 years ago