Pierre-Auguste Renoir's charge account uses the unpaid balance method to compute the

PLZ HELP!!! WILL MAKE BRAINLIEST!!!!!

KIM [24]

Answer:

C. 75 restaurant customers

Step-by-step explanation:

With the way the graph is laid out 75 restaurant customers seems to be the most logical answer.

7 0

3 years ago

Read 2 more answers

Name the quadrant in which the point (−7, 3) lie

gtnhenbr [62]

Answer:

The quadrant in which the point (−7, 3) lie is the 2nd qudrant

4 0

3 years ago

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In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 11% of voters a

Radda [10]

Answer:

a) 0.0214 = 2.14% probability that none of the people are Independent.

b) 0.8516 = 85.16% probability that fewer than 6 are Independent.

c) 0.8914 = 89.14% probability that more than 2 people are Independent.

Step-by-step explanation:

For each people, there are only two possible outcomes. Either they are independent, or they are not. For each person asked, the probability of them being Independent voters is the same. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

It is believed that 11% of voters are Independent.

This means that $p = 0.11$

A survey asked 33 people to identify themselves as Democrat, Republican, or Independent.

This means that $n = 33$

A. What is the probability that none of the people are Independent?

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

0.0214 = 2.14% probability that none of the people are Independent.

B. What is the probability that fewer than 6 are Independent?

This is

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X = 2) = C_{33,2}.(0.11)^{2}.(0.89)^{31} = 0.1724$

$P(X = 3) = C_{33,3}.(0.11)^{3}.(0.89)^{30} = 0.2202$

$P(X = 4) = C_{33,4}.(0.11)^{4}.(0.89)^{29} = 0.2041$

$P(X = 5) = C_{33,5}.(0.11)^{5}.(0.89)^{28} = 0.1463$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0214 + 0.0872 + 0.1724 + 0.2202 + 0.2041 + 0.1463 = 0.8516$

0.8516 = 85.16% probability that fewer than 6 are Independent.

C. What is the probability that more than 2 people are Independent?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X < 2) = 0.0214 + 0.0872 = 0.1086$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1086 = 0.8914$

0.8914 = 89.14% probability that more than 2 people are Independent.

8 0

3 years ago

Parte de Cálculo: En cada una de las siguientes proporciones, utilizar regla de tres para determinar el valor desconocido

sineoko [7]

Answer:

x = 4

x = 3

x = 10

x = 3

x = 16

x = 35

Step-by-step explanation:

$\frac{x}{5} :\frac{8}{10}$

x · 10 = 5 · 8

10x = 40

10x ÷ 10 = 40 ÷ 10

x = 4

$\frac{8}{12} :\frac{2}{x}$

x · 8 = 12 · 2

8x = 24

8x ÷ 8 = 24 ÷ 8

x = 3

$\frac{15}{3} :\frac{x}{2}$

x · 3 = 15 · 2

3x = 30

3x ÷ 3 = 30 ÷ 3

x = 10

$\frac{x}{6}:\frac{6}{12}$

x · 12 = 6 · 6

12x = 36

12x ÷ 12 = 36 ÷ 12

x = 3

$\frac{x}{8} :\frac{4}{2}$

x · 2 = 8 · 4

2x = 32

2x ÷ 2 = 32 ÷ 2

x = 16

$\frac{10}{2} :\frac{x}{7}$

x · 2 = 10 · 7

2x = 70

2x ÷ 2 = 70 ÷ 2

x = 35

5 0

3 years ago

Help please!!

ivanzaharov [21]

Answer:

answer is 100 pages per hour

5 0

3 years ago