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jarptica [38.1K]
3 years ago
14

The volume of a sphere is 562.51 in? How many inches are in the radius of the sphere?

Mathematics
2 answers:
dlinn [17]3 years ago
7 0

Answer:

C is the correct answer

Step-by-step explanation:

olya-2409 [2.1K]3 years ago
4 0

Answer:C

Step-by-step explanation:

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You have a 5-question multiple-choice test. Each question has four choices. You don’t know any of the answers. What is the exper
lana66690 [7]
2 out of 5 chance that you will attests get 3
4 0
3 years ago
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
Can someone explain to me how to answer questions like this? I always get confused!
seropon [69]

Answer:

(b) 5k+1

Step-by-step explanation:

https://www.symbolab.com/solver/step-by-step/simplify%20%5Cfrac%7B30k%5E%7B2%7D%2B6k%7D%7B5k%2Bk%7D

5 0
3 years ago
What IS 20 divided by 13 ?
goldenfox [79]

Solution,\mathrm{Convert\:improper\:fractions\:to\:mixed\:numbers}:\quad \frac{20}{13}=1\frac{7}{13}

Steps:

\frac{20}{13}

\mathrm{Long\:Division}\:\frac{20}{13}:\quad 1\quad \mathrm{Remainder}\quad \:7

Convert\:to\:mixed\:number:\:Quotient\frac{Remainder}{Divisor},\\\frac{20}{13}=1\frac{7}{13}

Hope\: This\: Helps!!!

4 0
3 years ago
Read 2 more answers
Calculamos la fuerza de atracción gravitatoria entre dos cuerpos de 12kg y 30kg separados 50cm
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Es 42kg porque cuando lo suma te da eso pruebalo 
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