Kim's Loan:
Let A = amount borrowed from bank A.
Let B = amount borrowed from bank B.
The total loan is $55,000.
Therefore
A + B = 55000 (1)
Bank A charges 8% interest. The interest after 1 year is 0.08A.
Bank B charges 11% interest. The interest after 1 year is 0.11B.
Total interest after 1 year is $5,000.
Therefore
0.08A + 0.11B = 5000 (2)
From (1), obtain
B = 55000 - A (3)
Substitute (3) into (2).
0.08A + 0.11(55000 - A) = 5000
0.08A + 6050 - 0.11A = 5000
-0.03A = -1050
A = -1050/-0.03 = 35000
Answer: Kim borrowed $35,000 from bank A.
Jack's Loan.
A = loan from bank A.
B = loan from bank B.
The total loan is $10,000.
Therefore
A + B = 10000 (4)
Bank A charges 5% interest and bank B charges 6% interest.
Total interest after 1 year is $530, therefore
0.05A + 0.06B = 530 (5)
From (4), obtain
B = 10000 - A (6)
Substitute (6) into (5).
0.05A + 0.06(10000 - A) = 530
0.05A + 600 - 0.06A = 530
-0.01A = -70
A = -70/-0.01 = 7000
Answer: Jack borrowed $7,000 from bank A.
1.06
1.501
1.506
1.605
Hope this helps
Answer:
2
Step-by-step explanation:
5+1+7+x+3+2+4=x+22
x+22 is divisible by 3,
multiple of 3 are 3,6,9,12,15,18,21,24,27,...
if x=2 ,then x+22=2+22=24
so smallest number is 2
Answer:
The standard deviation of the sampling distribution of the sample wait times is of 0.8 minutes.
Step-by-step explanation:
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30. Otherwise, the mean and the standard deviations holds, but the distribution will not be approximately normal.
Standard deviation 4 minutes.
This means that 
A sample of 25 wait times is randomly selected.
This means that 
What is the standard deviation of the sampling distribution of the sample wait times?

The standard deviation of the sampling distribution of the sample wait times is of 0.8 minutes.