Answer: A good estimation would be about 150.
Step-by-step explanation:
To estimate we can round instead of using exact numbers.
First day he saw about 30 birds and twice as many squirrels, meaning about 60 squirrels. This is about 90 total when added together.
Second day he saw 20 birds and about 40 squirrels so about 60 when added together.
Add day one and day two totals for...
90+60=150
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths.
Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday.
Jan - Friday
Feb - Monday
Mar - Monday
Apr - Thursday
May - Saturday
Jun - Tuesday
Jul - Thursday
Aug - Sunday
Sep - Wednesday
Oct - Friday
Nov - Monday
Dec - Wednesday
Now let's count how many times for each weekday, the 13th falls there.
Sunday - 1
Monday - 3
Tuesday - 1
Wednesday - 2
Thursday - 2
Friday - 2
Saturday - 1
The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc.
So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get:
Jan - Friday
Feb - Monday
Mar - Tuesday
Apr - Friday
May - Sunday
Jun - Wednesday
Jul - Friday
Aug - Monday
Sep - Thursday
Oct - Saturday
Nov - Tuesday
Dec - Thursday
And the weekday totals are:
Sunday - 1
Monday - 2
Tuesday - 2
Wednesday - 1
Thursday - 2
Friday - 3
Saturday - 1
And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year.
And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
Answer:
The answer is 1/4!
Step-by-step explanation:
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.
