Which unit rate is the lowest price per ounce? Choice A: 6 ounces of raisins for $2.98 Choice B: 13 ounces of raisins for $7.18

If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

Ilya [14]

This is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the "common difference" "d".

now, we know the 17th term is 34, let's go to the 15term then

34 + d
34 + d + d
34 + d + d + d
34 + d + d + d + d
34 + d + d + d + d + d
34 + d + d + d + d + d + d
34 + d + d + d + d + d + d + d
34 + d + d + d + d + d + d + d + d

now, notice, we got to the 15th term, which is 34 + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

$\bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}$

ok, now that we know what the common difference is, let's find the first term in the sequence using the 7th term of 34,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=5\\
a_7=34
\end{cases}
\\\\\\
a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5
\\\\\\
34=a_1+30\implies \boxed{4=a_1}$

and now let's use those 2 found folks, to get the 21st term,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=4\\
d=5\\
n=21
\end{cases}
\\\\\\
a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104$

3 0

3 years ago

Ayo runs a fairground game.In each turn, a player rolls a fair dice numbered 1 - 6 and spins a fair spinner numbered 1 - 12.It c

aksik [14]

Using the definition of expected value, it is found that Ayo can be expected to make a profit of £55.8.

The expected value is given by the sum of each outcome multiplied by it's respective probability.

In this problem:

The player wins $6, that is, Ayo loses £6, if he rolls a 6 and spins a 1, hence the probability is $\frac{1}{6} \times \frac{1}{12} = \frac{1}{72}$ .

The player wins $3, that is, Ayo loses £3, if he rolls a 3 on at least one of the spinner or the dice, hence, considering three cases(both and either the spinner of the dice), the probability is $\frac{1}{6} \times \frac{1}{12} + \frac{1}{6} \times \frac{11}{12} + \frac{5}{6} \times \frac{1}{12} = \frac{1 + 11 + 5}{72} = \frac{17}{72}$