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

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,

So, the weight c would be

Hence, the value of c would be 26.514 lb.

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah

answer:   x*94 +688*6   = 267*100

94x+4128=26700    

94x=22572     /94

x=240.128

step-by-step explanation: mark as brainlest reallly need it thx