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3 years ago

How to find missing number in equivalent ratio

Mathematics

1 answer:

Rashid [163]3 years ago

5 0

Explanation:

Consider ...

x/a = b/c . . . . . find x

Multiplying by the denominator under x gives ...

x = ab/c . . . . the value of the unknown.

____

In the case where the unknown is in the denominator, you can invert the ratios and solve as above:

a/x = c/b . . . . . note that x is in the denominator

x/a = b/c . . . . . equivalent equation, solve as above

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There were thirteen friends playing a video game online when six players quit.If each player left had eight lives how many lives

Lorico [155]

They had 56 lives in total

Explanation: if there were thirteen friends to begin with and six quit that would leave them with 7 friends. If those seven remaining friends each had 8 lives, you multiply those 7 friends by 8 lives which would get 56.

If all thirteen friends were still in the game then they would have 104 lives all together.

5 0

2 years ago

Read 2 more answers

A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c

olchik [2.2K]

The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in $E^{c}$ . Find $P(E^{c})$ .

Solution:

Part 1:

$E^{c}$ means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

$E^{c}=S-E \\ \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\ \\ 
E^{c}=(4,5,6)$

Thus the set compliment of E will contain the elements {4,5,6}.So

$E^{c}$ = {4,5,6}

Part 2)

$P(E^{c})$ means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

$P(E^{c})$ = (Number of Elements in $E^{c}$ )/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set $E^{c}$ = $n(E^{c})$ =3

So,

$P(E^{c})= \frac{n(E^{c}) }{n(S)} \\ \\ 
P(E^{c})= \frac{3}{12} \\ \\ 
P(E^{c})= \frac{1}{4}$

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3 years ago

Keely says that he’s glad that his morning coffee is sold in a monopolistically competitive market rather than a purely competit

aleksklad [387]

This suggests that he buys products with higher prices than other similar products. But this type of market(monopolistic competition) is great though, because there are lesser suppliers than in the pure competition where products are almost within the same price ranges. In a monopolistic competition, suppliers sell their products with a justifiable high price. This is also an advantage for the buyers, they would be able to consume and be satisfied with products of high quality.

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3 years ago

ASAP <br><br> AND HOW DID YOU GET THE ANSWER <br><br> ( WILL MARK BRIANIST)

andrew11 [14]

Answer:

a 17

Step-by-step explanation:

20 + 5 - 8

8 0

3 years ago

the ratio of peters money to henrys money is 4 : 3 there at first. after perter spent 12 dollars they had an equal amount of mon

MatroZZZ [7]

Using a system of equations, it is found that Peter had $48 at first.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Variable x: Peter's money.

Variable y: Henry's money.

The ratio of peters money to henrys money is 4 : 3, hence:

$\frac{x}{y} = \frac{4}{3}$

After Peter spent $12, they had the same amount, hence:

y = x - 12.

Then, replacing in the ratio:

$\frac{x}{y} = \frac{4}{3}$

$\frac{x}{x - 12} = \frac{4}{3}$

4(x - 12) = 3x

x = 48.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

4 0

2 years ago