. A shipyard makes a container ship that can withstand the total amount of weight W, which is normally distributed with mean of 600 tons and standard deviation of 60 tons. Let us assume that the weight of a single container that will be loaded to the container ship is also normally distributed with mean of 4 tons and standard deviation of 0.4 tons. What is the maximum number of containers that the ship can load and still have at least a 90% chance to not exceed its weight limit

the maximum number of containers that the ship can load is 170

Step-by-step explanation:

Given the data in the question;

W ~ N( 600, 60 )

S ~ N( 4n, 0.4√n )

so

p( W > S ) = 0.90

⇒ P( W - S > 0) = 0.9 ------ let this be equation 1

now, since W and S are independent

Mean( W - S ) = Mean( W ) - Mean( S 0 = 600 - 4n

and

SD( W - S ) = √( var(W) + var(S) ) = √( 60² + 0.4²n)

hence;

W - S ~ N( 600 - 4n, √( 60² + 0.4²n) )

now, from equation one, P( W - S > 0) = 0.9

from z- table

------------------ let this be equation 2

now, we square both sides of equation 2

we cross multiply

16n² + 360000 - 4800n = 1.643524( 3600 + 0.16n )

16n² + 360000 - 4800n = 5916.6864 + 0.26296384n

16n² + 360000 - 5916.6864 - 4800n - 0.26296384n = 0

16n² + 354083.3136 - 4800.26296384n = 0    

16n² - 4800.26296384n + 354083.3136 = 0  

solving the quadratic equation, we know that;

x = -b±√( b² - 4ac ) / 2a

so we substitute

x = [-(-4800.26296384) ±√( (-4800.26296384)² - (4 × 16 × 354083.3136)] / [2×16]

x = [ 4800.26296384 ±√( 23042524.522 - 22661332.0704 ] / 32

x = [ 4800.26296384 ±√(381192.4516) ] / 32

x = [ 4800.26296384 ± 617.4078 ] / 32

Hence;

x = [ 4800.26296384 - 617.4078 ] / 32 or  [ 4800.26296384 + 617.4078 ] / 32

x = 131   or  170

Therefore, the maximum number of containers that the ship can load is 170

35 is an outlier

step-by-step explanation:

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240