Answer: The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is 0.596
Step-by-step explanation:
Since the weights of catfish are assumed to be normally distributed,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = weights of catfish.
µ = mean weight
σ = standard deviation
From the information given,
µ = 3.2 pounds
σ = 0.8 pound
The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is is expressed as
P(x ≤ 3 ≤ 5.4)
For x = 3
z = (3 - 3.2)/0.8 = - 0.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.401
For x = 5.4
z = (5.4 - 3.2)/0.8 = 2.75
Looking at the normal distribution table, the probability corresponding to the z score is 0.997
Therefore,.
P(x ≤ 3 ≤ 5.4) = 0.997 - 0.401 = 0.596
28/8 or 7/2 is the answer to that
Answer:
SAS
Step-by-step explanation:
In triangle ABC and triangle EFD
1. BC = CD (S) Given
2. < BCD = < EFD (A) being vertically opposite. angles.
3. AC = CE (S) Given
Hence
By SAS postulate Triangle ABC and triangle EFD are congruent.
HOPE IT HELPS :)❤
