The calendar obviously has an integral number of years and months in 400 years. If it has an integral number of weeks, then it will repeat itself after that time. The rules of the calendar eliminate a leap year in 3 out of the four century years, so there are 97 leap years in 400 years. The number of excess days of the week in 400 years can be found by ...
(303·365) mod 7 + (97·366) mod 7 = (2·1 + 6·2) mod 7 = 14 mod 7 = 0
Thus, there are also an integral number of weeks in 400 years.
The first day of the week is the same at the start of every 400-year interval, so the calendar repeats every 400 years.
Answer:
y < - 12
Step-by-step explanation:
y + 15 < 3 ( subtract 15 from both sides )
y < - 12
Answer:
Step-by-step explanation:
B.(10,1)
50*.6=30
so if you solve this equation it would be 30 dogs in the shelter
We can find critical value by using t - table.
For using t - table we need degree of freedom and alpha either for two tailed test or one tailed test.
We can determine degree of freedom by subtracting sample size from one.
So in given question sample size is 23. So we can say degree of freedom(df) for sample size 23 is
df = 23 - 1= 22
Now we have to go on row for degree of freedom 22.
After that we need to find alpha either for two tailed test or one tailedl test.
Confidence level is 99%. We can convert it into decimal as 0.99.
So alpha for two tailed test is 100 - 0.99 = 0.01
Alpha for one tailed test is 0.01/2 = 0.005.
So we will go on column for 0.01 for two tailed test alpha or 0.005 for one tailed test alpha.
SO the critical value 22 degree of freedom and 0.01 two tailed alpha is 2.819 from t - table.
