Eight plus the quotient of a number and 3 is -2

The number of measles cases increased 13.6 % to 57 cases this year, what was the number of cases prior to the increase? (Express

ELEN [110]

Answer:

The number of cases prior to the increase is 50.

Step-by-step explanation:

It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.

We need to find the number of cases prior to the increase.

Let x be the number of cases prior to the increase.

x + 13.6% of x = 57

$x+x\times \frac{13.6}{100}=57$

$x+0.136x=57$

$1.136x=57$

Divide both the sides by 1.136.

$\frac{1.136x}{1.136}=\frac{57}{1.136}$

$x=50.176$

$x\approx 50$

Therefore the number of cases prior to the increase is 50.

7 0

3 years ago

An epidemiologist is observing the decay pattern of a pathogenic bacteria after applying a vaccine. He starts with 1,000 bacteri

Zielflug [23.3K]

Answer:

8%=1 hour

?=48 hours

48×8

384

1000-384

616 bacteria

6 0

4 years ago

Read 2 more answers

Whats 30 times 30 plus 45 minus 456

aleksandr82 [10.1K]

30 times 30 plus 45 minus 456 is 489

4 0

3 years ago

Read 2 more answers

Name all of the radii of the circle. Ok guys im soooo sorry!!!

stich3 [128]

Answer: B

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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 $3 \sim 2$ .

3 0

3 years ago