Answer:
2.) $520
3.) less
Step-by-step explanation:
I'm going to assume that the interst is compoudning and is convertable once a year
The compound interest formula for interest compounding only once a year is as follows
plug in the numbers and get
3.) If the interest rate is lower at the credit union he would obvioulsy be paying less (assuming that this interest rate is convertable annually as well)
Answer:
9 Pizzas.
Step-by-step explanation:
Since Sam has $150 in total, and already spent $40 on drinks, we need to deduct that from the total ($150 - $40), in turn leaving us with $110. Since we aren't sure of the amount of pizzas bought yet, lets use the variable x in its place. Since each pizza costs $12, we need to use $12(x) = $110, $110 again being the total after Sam bought the drinks. After dividing $110 by $12, we are left with 9 pizzas, with $2 extra. Therefore, he can buy 9 pizzas.
The answer must have to be 40
We have the equations:
<span>C(n,k) / C(n,k+1) = 1/2
</span>C(n,k+1) / C(n, k+2) = 2/3<span>
</span>[n! / (n-k)!k! ]/[ <span>n! / (n- (k+1))! (k+1)! ]= 1/2
You can cancel out similar terms. The definition of factorials is this
n! = n(n -1)(n -2)(n -3)...1
So,
</span> (n- (k+1))! (k+1)! / (n-k)!k! = 1/2
(n- k - 1)(n - k - 1 -1)! (k+1)(k + 1 - 1)! / (n-k)!k! = 1/2
(n- k - 1)(n - k - 2)! (k+1)(k)! / (n-k)!k! = 1/2
Cancel out terms.
(n- k - 1)(n - k - 2)! (k+1)/ (n-k)!= 1/2 [eqn 1]
[n! / (n-k+1)!(k+1)! ]/[ n! / (n- (k+2))! (k+2)! ]= 2/3
(n- (k+2))! (k+2)! / (n-(k+1))!(k+1)! = 2/3
(n- k-2)! (k+2)! / (n-k-1)!(k+1)! = 2/3
(n- k-2)(n - k -3)! (k+2)( k+1)k! / (n-k-1)(n-k-2)(n-k-3)!(k+1)k! = 2/3
Cancel out terms
(k+2)/ (n-k-1) = 2/3
You can solve for n in terms of k and substitute this to the fist equation which will allow you to solve for k.