19, 38, 76, ... find the 8th term

Molly opened a savings account with a one‐time deposit of $100 that will be left in the account for at least 5 years. The saving

Anettt [7]

The amount of interest Molly will earn after 5 years on a deposit of $\$100$ compounded annually over 5 years is $\$27.63$

First, we need to find the future value of her investment, then we subtract the original deposit from it to get the amount of interest she will get at the end of 5 years.

The future value of an investment that is compounded annually is given by

$A=P(1+r)^t$

where

$A=\text{Amount of money in Molly's account after 5 years}\\P=\text{Molly's initial deposit into her account}=\$500\\r=\text{The annual interest rate as a decimal}=0.05\\t=\text{The time the money is invested (in years)}=5$

Substituting the available values into the formula and solving

$A=100(1+0.05)^5=100(1.05)^5 \approx \$ 127.63$

The interest Molly will earn after 5 years is

$A-P=\$127.63-\$100=\$27.63$

Therefore, the amount of interest Molly will earn after 5 years on a deposit of $\$100$ compounded annually over 5 years is $\$27.63$

Learn more about compound interest here: brainly.com/question/21270833

8 0

3 years ago

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

kozerog [31]

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$ .

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$ , $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$ . However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$ . For any two elements $a,\, b \in S$ , $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$ .

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

Reflexivity: for any $a \in S$ , the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$ .)

Symmetry: for any $a,\, b \in S$ , $a \sim b$ if and only if $b \sim a$ . In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$ , or neither is in $R\!$ .

Transitivity: for any $a,\, b,\, c \in S$ , if $a \sim b$ and $b \sim c$ , then $a \sim c$ . In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$ , then $(a,\, c)$ also needs to be in $R\!$ .

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$ ) in this question is indeed reflexive. $(1,\, 1)$ , $(2,\, 2)$ , and $(3,\, 3)$ (one pair for each element of $S$ ) are all elements of $R\!$ .

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$ .

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$ . However, $(3,\, 2) \not \in R$ . In other words, under relation $R\!$ , $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$ .