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ivann1987 [24]
2 years ago
14

Gavin made a wooden planter in the shape of a cube. He measured the inside of the planter and found that each side is 38.75 cm l

ong. To the nearest hundredth, how much dirt does Gavin need to completely fill his planter?
Mathematics
2 answers:
notka56 [123]2 years ago
7 0

Answer:

3/5

Step-by-step explanation:

iogann1982 [59]2 years ago
4 0

Answer:

58,185.55

Step-by-step explanation:

You might be interested in
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
2 years ago
1. Find the measurement of angle J.<br><br> 2. (5x - 13) (3x + 17)
Alex Ar [27]

Answer:

Q1. x= 18, y=59

Q2. m∠J= 56°

Step-by-step explanation:

Q1. (3x +5)°= y° (base ∠s of isos. △)

y= 3x +5 -----(1)

(3x +5)° +y° +(4x -10)°= 180° (∠ sum of △)

3x +5 +y +4x -10= 180

7x +y -5= 180

7x +y= 180 +5

7x +y= 185 -----(2)

Substitute (1) into (2):

7x +3x +5= 185

10x= 185 -5

10x= 180

x= 180 ÷10

x= 18

Substitute x= 18 into (1):

y= 3(18) +5

y= 59

Q2. (5x -13)°= (3x +17)° (base ∠s of isos. △)

5x -13= 3x +17

5x -3x= 17 +13

2x= 30

x= 30 ÷2

x= 15

∠LKJ

= 3(15) +17

= 62°

∠KLJ= 62° (base ∠s of isos. △)

m∠J

= 180° -62° -62° (∠ sum of △JKL)

= 56°

8 0
2 years ago
Help please help me
BARSIC [14]

Step-by-step explanation:

if you're looking for x

x=180-42-51= 87°

and if you're looking for D

D= 180-42= 138°

7 0
3 years ago
Can someone help me !
castortr0y [4]

Answer:

4 x 4 is 12

3 x 2 is 6

12 plus 6 is 18

Answer is 18

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Find two consecutive odd integers whose product is 1443
liberstina [14]

Let's x represents the first odd integer

The next consecutive integer would be represented as x+2.

So x and x+2 being multiplied together will give us 1443:

(x)(x+2)=1443

x^2+2x=1443

x^2+2x-1443=0

By solving the quadratic equation using the quadratic formula, x will equal to 37 and -39.

Since the problem says "integers", so I'm assuming two pairs of consecutive integers would be ok.

With that said you 1st pair will be: 37,39

Your second pair will be: -37,-39.



6 0
2 years ago
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