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Gelneren [198K]
3 years ago
11

The weight distribution of parcels sent in a certain manner is normal with mean value 17 lb and standard deviation 3.3 lb. The p

arcel service wishes to establish a weight value c beyond which there will be a surcharge. What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight? (Round your answer to three decimal places.) c = lb
Mathematics
1 answer:
Yakvenalex [24]3 years ago
7 0

Answer: The value of c would be 26.514 lb.

Step-by-step explanation:

Since we have given that

Mean = 17 lb

Standard deviation = 3.3 lb

At 99%  level of significance, z = 2.58

So, it becomes,

Z=\dfrac{X-\mu}{\sigma}\\\\2.58=\dfrac{X-17}{3.3}\\\\2.58\times 3.3=X-17\\\\8.514+17=X\\\\X=25.514

So, the weight c would be

c=25.514+1\\\\c=26.514

Hence, the value of c would be 26.514 lb.

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nalin [4]

Ans(1):

Given equation is f(x)=-1.5x+6

we can plug any number like x=0 and x=2 to find the f(x) also called y-value

plug x=0

f(x)=-1.5x+6 =-1.5*0+6 =0+6 =6

Hence first point is (0,6)

plug x=2

f(x)=-1.5x+6 =-1.5*2+6 =-3+6 =3

Hence first point is (2,3)

now we can graph both points then join them to get final graph of f(x)=-1.5x+6

---------------------

Ans(2):

We can repeat exactly same process for f(x) = -1/2x-5.

So the final graph will look like attached picture:


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3 years ago
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hram777 [196]

Answer:

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Domain is -3 \leq x \leq 2.

Range is -5 \leq y \leq 4

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Step-by-step explanation:

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