The answer is C to this question
Answer:
50
Step-by-step explanation:
Sum Even numbers
n = 50
d = 2
a1 = 2
The last number is
an = a1 + (n-1)d
an = 2 + (50 - 1)*2
an = 2 + 49 * 2
an = 2 + 98
an = 100
Sum of the even numbers
Sum = (a1 + a50)*n/ 2
Sum = (2 + 100)*50/2
sum = 102 * 25
sum = 2550
Sum of the first 50 odd numbers
a1 = 1
n = 50
d = 2
l = ?
Find l
l = a1 + (n - 1)*2
l = 1 + 49*2
l = 99
Sum
Sum = (1 + 99)*50/2
Sum = 2500
The difference and answer is 2550 - 2500 = 50
1-2. The best estimate for the population mean would be sample mean of 60 gallons. Since we know that the sample mean is the best point of estimate. Since sample size n=16 is less than 25, we use the t distribution. Assume population from normal distribution.
3. Given a=0.1, the t (0.05, df = n – 1 = 15)=1.75
4. xbar ± t*s/vn = 60 ± 1.75*20/4 = ( 51.25, 68.75)
5. Since the interval include 63, it is reasonable.
I think the answer would be B that’s the only one I consider
