I think it's 21% when rounded to a whole number
Answer: 912
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Work Shown:
The starting term is a1 = 3. The common difference is d = 5 (since we add 5 to each term to get the next term). The nth term formula is
an = a1+d(n-1)
an = 3+5(n-1)
an = 3+5n-5
an = 5n-2
Plug n = 19 into the formula to find the 19th term
an = 5n-2
a19 = 5*19-2
a19 = 95-2
a19 = 93
Add the first and nineteenth terms (a1 = 3 and a19 = 93) to get a1+a19 = 3+93 = 96
Multiply this by n/2 = 19/2 = 9.5 to get the final answer
96*9.5 = 912
I used the formula
Sn = (n/2)*(a1 + an)
where you add the first term (a1) to the nth term (an), then multiply by n/2
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As a check, here are the 19 terms listed out and added up. We get 912 like expected.
3+8+13+18 +23+28+33+38 +43+48+53+58 +63+68+73+78 +83+88+93 = 912
There are 19 values being added up in that equation above. I used spaces to help group the values (groups of four; except the last group which is 3 values) so it's a bit more readable.
Answer: disagree its backwards
Step-by-step explanation: the x intercept always comes first so its actually (-4,2)
Answer:
4032 different tickets are possible.
Step-by-step explanation:
Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.
To find : How many different tickets are possible ?
Solution :
In the first race there are 9 ways to pick the winner for first and second place.
Number of ways for first place - 
Number of ways for second place - 
In the second race there are 8 ways to pick the winner for first and second place.
Number of ways for first place -
Number of ways for second place - 
Total number of different tickets are possible is
Therefore, 4032 different tickets are possible.
