Answer: the probability that a randomly selected Canadian baby is a large baby is 0.19
Step-by-step explanation:
Since the birth weights of babies born in Canada is assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = birth weights of babies
µ = mean weight
σ = standard deviation
From the information given,
µ = 3500 grams
σ = 560 grams
We want to find the probability or that a randomly selected Canadian baby is a large baby(weighs more than 4000 grams). It is expressed as
P(x > 4000) = 1 - P(x ≤ 4000)
For x = 4000,
z = (4000 - 3500)/560 = 0.89
Looking at the normal distribution table, the probability corresponding to the z score is 0.81
P(x > 4000) = 1 - 0.81 = 0.19
Literally just add it all up.
312.8+177.2+209=699 miles
Answer:
Step-by-step explanation:
Answer: 63lbs
Step-by-step explanation:
The truck can only hold 16crates and each crate weigh 12lbs, the total weight of the 16 crates will be 12×16= 192lbs
This shows that the truck can only contain extra load of 1200lbs - 192lbs = 1008lbs (excluding weight of crates).
To get the shipment weigh close to the total of 1200lbs, the truck must be loaded with engine components not more than 1008lbs.
Since we have 16 crates to fill with engine components not more than 1008lb, each crates will therefore must not exceed 1008/16 pounds of engine components which is equivalent to 63lbs.
Therefore, the manager should instruct the workers to put 63lbs of machine components in "each crate" in order to get the shipment weight as close as possible to 1200 lbs.
Answer:
9. 
11. 
Step-by-step explanation:
9)
11)
Hope this helps!
