Quick answer this question whoever gives correct answer gets brainliest

Mathematics

1 answer:

liq [111]3 years ago

8 0

Answer:

50 envelopes for 2.49

Step-by-step explanation:

2.49/50= .05

5.50/90= .06

the first per is less than the 90 envelopes one therefore the answer is 50 for 2.49

You might be interested in

Solve 4(5x - 2) = 2(9x + 3)

saveliy_v [14]

X will equal 7.
4(5x-2)=2(9x+3)
20x-8=18x+6
20x-18x=8+6
2x=14
x=7

7 0

3 years ago

Read 2 more answers

{2x + y = 8 x + -4y = 5

bonufazy [111]

Answer:

x 29/7

y -2/7

Step-by-step explanation:

4 0

3 years ago

Abby, Deborah, Mei-Ling, Sam, and Roberto work in a firm's public relations office. Their employer must choose two of them to at

sattari [20]

Answer:

a) Ω = { {Abby, Deborath}, {Abby, Mei-Ling}, {Abby, Sam}, {Abby,Roberto}, {Deborah, Mei-Ling}, {Deborah, Sam}, {Deborah, Roberto}, {Mei-Ling, Sam}, {Mei-Ling, Roberto}, {Sam, Roberto} }

b) 0.1

c) 0.4

d) 0.3

Step-by-step explanation:

a) The sample space must contain every possible combination of two names. Since we dont care about the order of the chosen names, we can describe every element of the sample space as a subset of 2 elements of the set {Abby, Deborah, Mei-Ling, Sam, Roberto}. That subset will represent the names of the chosen persons. With this in mind, we conclude that the sample space is

Ω = { {Abby, Deborath}, {Abby, Mei-Ling}, {Abby, Sam}, {Abby,Roberto}, {Deborah, Mei-Ling}, {Deborah, Sam}, {Deborah, Roberto}, {Mei-Ling, Sam}, {Mei-Ling, Roberto}, {Sam, Roberto} }

b) The cardinality of the sample space Ω is 10. Since all choices are equally likely, any choice will have probability $\frac{1}{10} = 0.1$ , because the 10 of them combined must add up 1.

c) We need to find all possible choices that includes Mei-Ling, those will be our favourable cases. The amount of favourable cases must be divided to the total amount of cases (the cardinality of Ω) in order to obtain the probability of Mei-Ling being chosen. Mei-Ling is included on 4 choices (one for each of her partners), this means that she has a probability of $\frac{4}{10} = 0.4$ to being chosen.

d) We have 3 favourable cases ,the choices {Abby, Deborah}, {Abby, Mei-Ling} and {Deborah, Mei-Ling}, which neither of them contain a man. By dividing that number to the total number of cases, we obtain a probability of $\frac{3}{10} = 0.3$ that neither of the two men are chosen