Answer:
No solution
Step-by-step explanation:
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by 2 gives the next term. In other words, <em>a n +a1 ⋅ r n -1</em><em> </em>
Geometric Sequence:<em> </em><em>r = 2</em>
The series given has a value of<em> </em><em>r </em>such that <em>r > 1 or
r < − 1.</em> Therefore, the infinite sum cannot be calculated.<em>
</em>
<em>
</em>
<em>
</em>
A regular polygon has about 5 lines of symmetry. The number of lines of symmetry is equivalent to the number of sides.
Y = total amount of money spent
x = cost of one drink
So our expression is:
4x + 12.75
4x stands for 4 drinks times x price of one drink
The 12.75 is a one time fee for the pizza, since Kiyo isn't buying multiple pizzas.
The problem also asks us to find the total price, so we write an equation:

We insert what x is which in this case is $3

Order of Operations tells us to multiply first:

Now add your values:

Therefore, Kiyo spent a total amount of 24.75 at Paola's pizza.
Let's check our problem just in case:

Since we now know what y is equal to, we can plug that into our original equation and check just to verify
We should subtract 12. on both sides of the equation because of inverse operations:

We are left with:

We divide:

Our result is:

Which checks out because each drink was in fact, $3 dollars as stated in the word problem. Hope that was thorough enough.
Answer:

Step-by-step explanation:
Given
Let
Undergraduates
Graduates
So, we have:
-- Total students
--- students to select
Required

From the question, we understand that 2 undergraduates are to be selected; This means that 2 graduates are to be selected.
First, we calculate the total possible selection (using combination)

So, we have:





Using a calculator, we have:

The number of ways of selecting 2 from 3 undergraduates is:




The number of ways of selecting 2 from 5 graduates is:




So, the probability is:




Step-by-step explanation:
How are we supposed to help...?