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3 years ago

Weekly wages at a certain factory are

Mathematics

2 answers:

Snezhnost [94]3 years ago

6 0

Answer: 34

Step-by-step explanation:

Hope this helps!

Naya [18.7K]3 years ago

4 0

Answer:

Step-by-step explanation:

Weekly wages at a certain factory are

normally distributed. The formula for normal distribution is expressed as

z= (x - u)/s

Where

u = mean

s = standard deviation

x = weekly wages

From the given information,

u = 400

s = 50

The probability that a worker

selected at random makes between

$350 and $400 is expressed as

P(350 lesser than or equal to x lesser than or equal to 400)

For x = 350

z = (350 - 400)/50 = -50/50 = -1

z = -1

From the normal distribution table, the corresponding z score is 0.1587

For x = 400

z = (400 - 400)/50 = 0/50 = 0

z = 0

From the normal distribution table, the corresponding z score is 0.5

P(350 lesser than or equal to x lesser than or equal to 400)

= 0.5 - 0.1587 = 0.3413

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Read 2 more answers

An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d

denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. $w\times 2w\times h=36$

$h=\frac{18}{w^2}$

The equation form when top is open,

$f(w)=2w^2+2wh+2(2w)h$

Substitute the value of h,

$f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})$

$f(w)=2w^2+\frac{36}{w}+\frac{72}{w}$

$f(w)=2w^2+\frac{108}{w}$

Derivate w.r.t 'w',

$f'(w)=4w-\frac{108}{w^2}$

For critical point put it to zero,

$4w-\frac{108}{w^2}=0$

$4w=\frac{108}{w^2}$

$w^3=27$

$w^3=3^3$

$w=3$

Derivate the function again w.r.t 'w',

$f''(w)=4+\frac{216}{w^3}$

For w=3, $f''(3)=4+\frac{216}{3^3}=12 >0$

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = $\frac{18}{3^2}=2\ ft$

Therefore, the dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

4 0

4 years ago