7 plus 7 equals 14. Add one to that and you have 15 which is the answer.
Answer: The open interval would be (31.4,42.5).
Step-by-step explanation:
Since we have given that
mean = 36.9
Standard deviation = 16.5
n = 48
At 98% confidence interval, z = 2.33
So, Interval would be
![\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%5Cpm%20z%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%5C%5C%3D36.9%5Cpm%202.33%5Cdfrac%7B16.5%7D%7B%5Csqrt%7B48%7D%7D%5C%5C%5C%5C%3D36.9%5Cpm%205.549%5C%5C%5C%5C%3D%2836.9-5.5%2C36.9%2B5.6%5C%5C%5C%5C%3D%2831.4%2C42.5%29)
Hence, the open interval would be (31.4,42.5).
Answer:
33,600
Step-by-step explanation:
length=20m
wide=14m
area= 28m²
1m= 28
cost = 120*28= rs33600 per m²
Supplementary would be like 6&7 or 2&3 and vertical would be like 6&8 or 7&5
Answer:
There are 4 possibilities
Step-by-step explanation
Each customer has 2 ways to receive the money. For each way, the other customer has 2 option, giving a total of 2*2 = 4. Here are the possibilities, separating in 2 different cases.
Case 1: customer 1 withdraws 2 $20 bills
option 1.1: customer 1 withdraws 2 $20 bills AND customer 2 withdraws 2 $20 bills
option 1.2: customer 1 withdraws 2 $20 bills AND customer 2 withdraws 4 $10 bills
Case 2: customer 1 withdraws 4 $10 bills
option 2.1: customer 1 withdraws 4 $10 bills AND customer 2 withdraws 2 $20 bills
option 2.2: customer 1 withdraws 4 $10 bills AND customer 2 withdraws 4 $10 bills
options 1.1, 1.2, 2.1 and 2.2 give us a total of four ways for the customers to retrieve thier money.