Given
mean of 406 grams and a standard deviation of 27 grams.
Find
The heaviest 14% of fruits weigh more than how many grams?
Explanation
given
mean = 406 gms
standard deviation = 27 gms
using standard normal table ,
![\begin{gathered} P(Z>z)=14\% \\ 1-P(Zso , [tex]\begin{gathered} x=z\times\sigma+\mu \\ x=1.08\times27+406 \\ x=435.16 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%28Z%3Ez%29%3D14%5C%25%20%5C%5C%201-P%28Zso%20%2C%20%5Btex%5D%5Cbegin%7Bgathered%7D%20x%3Dz%5Ctimes%5Csigma%2B%5Cmu%20%5C%5C%20x%3D1.08%5Ctimes27%2B406%20%5C%5C%20x%3D435.16%20%5Cend%7Bgathered%7D)
Final Answer
Therefore , The heaviest 14% of fruits weigh more than 435.16 gms
Answer: A = π r2
r=8 and 8 squared is 64 and 64×3.14=200.96 or rounded to the nearest whole number it would be 201
Step-by-step explanation: brainliest please?
Well the answer is -2y-4 but the different terms in this problem are 4 and 5y and 3y where 5y and 3y are like terms
Add 1
0.7+1=1.7
1.7+1= 2.7
etc., etc.
Answer:
The percentage of overdue accounts are held by customers in the "risky credit" category is 62.5%
Step-by-step explanation:
Customers in the "risky" category (25% of total accounts) allow their accounts to go overdue 50% of the time on average.
That means that on average, 12.5% of total accounts is overdue.
0.25*0.50 = 0.125
In the "good credit" category only 10% goes overdue. That means 7,5% of total accounts goes overdue in this category.
0.75*0.10=0.075
The total accounts that go overdue is 0.125+0.075 = 0.200.
The percentage of overdue accounts held by customers in the "risky credit" category is:
0.125/0.200 = 0.625 or 62.5%
