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LiRa [457]
3 years ago
14

You are responsible for insuring that 5550 finished goods are delivered to the new store in Springfield, MO. However, you are al

so aware of the following statistics: Expected theft within the warehouse prior to shipping: 0.5% Expected damages as a result of transportation: 1.7% How many finished goods should you begin with so that, even with the theft and damages, you will have at least 5550 items in Springfield, MO?! Your Answer:
Mathematics
1 answer:
Kamila [148]3 years ago
3 0

Answer:

5673 finished goods

Step-by-step explanation:

Given:

Number of finished goods required = 5550

Expected percentage theft within the warehouse prior to shipping = 0.5%

Expected percentage damages as a result of transportation = 1.7%

Now,

Number of Expected theft within the warehouse prior to shipping

= \frac{0.5}{100}\times5550

= 27.75 units ≈ 28

and,

Number of Expected damages as a result of transportation

= \frac{1.7}{100}\times5550

= 94.35 units ≈ 95

thus,

number of units of goods to be extra to begin with = 28 + 95

= 123 units

Hence,

there should be 5550 + 123 = 5673 finished goods so as to counter the theft and damages.

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Answer:

Option A  - ∑Fx and ∑τz and ∑Fy

Step-by-step explanation:

All the force will be in x and y plane only

So Torque will be in z plane

These 3 quantities should be 0

∑Fx and ∑τz and ∑Fy

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4 years ago
An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had
Stolb23 [73]

Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2  CI 95 %  =  (  0,056 ;  0,164 )

Step-by-step explanation:

Sample information for people under 18

n₁  =  500

x₁ =  180

p₁  =  180/ 500    p₁  =  0,36    then  q₁  =  1 -  p₁     q₁ =  0,64

Sample information for people over 18

n₂  =  600

x₂  =  150

p₂  =  150 / 600   p₂ =  0,25   then   q₂  =  1 - p₂   q₂ =  1 - 0,25   q₂ = 0,75

Hypothesis Test

Null hypothesis                        H₀              p₁  =  p₂

Alternative Hypothesis           Hₐ              p₁  ≠  p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

Testin at CI =  95 %    significance level is  α = 5 %   α  =  0,05  and

α/ 2  =  0,025   z (c) for that α  is from z-table:

z(c) = 1,96

To calculate   z(s)

z(s)  =  ( p₁  -   p₂ ) / EED

EED = √(p₁*q₁)n₁  +  (p₂*q₂)/n₂

EED = √( 0,36*0.64)/500  +  (0,25*0,75)/600

EED = √0,00046  +  0,0003125

EED = 0,028

( p₁  -  p₂  )  =  0,36  -  0,25  = 0,11

Then

z(s)  =  0,11 / 0,028

z(s) = 3,93

Comparing  z(s) and  z (c)    z(s) > z(c)

z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the idea of equals means

Q2  CI  95 %   =  (  p₁  -  p₂  ) ±  z(c) * EED

CI 95%  =  ( 0,11   ±  1,96 * 0,028 )

CI 95%  = (  0,11  ±  0,054 )

CI 95 %  =  (  0,056 ;  0,164 )

8 0
3 years ago
If f(x)=2x+5x and g(x)=3x-5,find (f+g)(x)
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The answer is (f+g)(x)=10x-5

Step-by-step explanation:

Given that f(x)=2x+5x and g(x)=3x-5

To find : (f+g)(x)

Now, assign the value g(x)=3x-5 in (f+g)(x)

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Also, assign the value f(x)=2x+5x in (f+g)(x)

(f+g)(x)=(2x+5x+3x-5)

Now, adding the like terms , we get,

(f+g)(x)=10x-5

Thus, the value of (f+g)(x) is (f+g)(x)=10x-5

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4 years ago
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Answer:

32 school hrs in a week.

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3 years ago
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