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4 years ago

15.) A ticket to last years homecoming dance was $12.50. If the total revenue from ticket sales was

Mathematics

1 answer:

Arisa [49]4 years ago

6 0

Answer:

1527

Step-by-step explanation:

just do 19,087.50 divided by 12.50

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Please help! Im stuck on this question

KonstantinChe [14]

Answer:

(x, y) = (-9, -1)

Step-by-step explanation:

Start by graphing two equations on the graph through the y-intercept and the slope.

Since they have different slope, they would intercept somewhere on the graph, which you can see clearly where they intercept from the drawing I included below.

*sorry for the terrible drawing

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3 years ago

Please please please help!!!

andreyandreev [35.5K]

The answer would be 6, hope this helps

4 0

3 years ago

Read 2 more answers

A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes a) are there in total?

attashe74 [19]

Answer:

There are 256 ways in total.

There are 56 possible outcomes contain exactly three heads.

The possible outcomes contain at least three heads is 219.

The possible outcomes contain the same number of heads and tails are 70.

Step-by-step explanation:

Consider the provided information.

A coin is flipped eight times where each flip comes up either heads or tails.

Part (a) How many possible outcomes are there in total?

Each time we flip a coin it comes up either heads or tail.

Therefore the total number of ways are:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8=256$

Hence, there are 256 ways in total.

Part (b) contain exactly three heads?

We want exactly 3 heads, therefore,

n=8 and r=3

According to the definition of combination: $\binom{n}{r}=\frac{n!}{r!(n-r)!}$

$\binom{8}{3}=\frac{8!}{3!(5)!}=56$

Hence, there are 56 possible outcomes contain exactly three heads.

Part (c) contain at least three heads?

For 3 heads: $\binom{8}{3}=\frac{8!}{3!(5)!}=56$

For 4 heads: $\binom{8}{4}=\frac{8!}{4!(4)!}=70$

For 5 heads: $\binom{8}{5}=\frac{8!}{5!(3)!}=56$

For 6 heads: $\binom{8}{6}=\frac{8!}{6!(2)!}=28$

For 7 heads: $\binom{8}{7}=\frac{8!}{7!(1)!}=8$

For 8 heads: $\binom{8}{8}=1$

Now add them as shown:

56+70+56+28+8+1=219

Hence, the possible outcomes contain at least three heads is 219.

Part (d) contain the same number of heads and tails?

Same number of heads and tails means that the value of r=4.

Therefore,

$\binom{8}{4}=\frac{8!}{4!(4)!}=70$

Hence, the possible outcomes contain the same number of heads and tails are 70.

6 0

3 years ago

A bag contains 240 marbles that are either red, blue, or green. The ratio of red to blue to green marbles is 5:2:1. If one-third

Andrej [43]

The fraction of the remaining marbles in the bag will be blue is 6/17.

<h3>What fraction of the remaining marbles in the bag will be blue?</h3>

The first step is to determine the initial number of marbles in the bag:

Initial number of red marbles in the bag : (5/8) x 240 = 150

Initial number of blue marbles in the bag : (2/8) x 240 = 60

Initial number of green marbles in the bag : 240 - 150 - 60 = 30

Number of red marbles remaining after 1/3 is removed = (1 - 1/3) x 150

2/3 x 150 = 100

Number of green marbles remaining after 2/3 is removed = (1 - 2/3) x 30

1/3 x 30 = 10

Total number of marbles now in the bag : 100 + 10 + 60 = 170

Fraction of blue marbles = 60 / 170 = 6/17

To learn more about ratios, please check: brainly.com/question/9194979

#SPJ1

7 0

2 years ago

What would you do with "X" and "y"<br> using the substitution method?

maw [93]

Answer:

They are used in place fir an unknow value

Step-by-step explanation:

7 0

3 years ago