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2 years ago

15

Mr. Spencer sells furniture. He earned $225 for selling 5 sofas. Mr. Spencer earned $300 for selling 8 sofas. Which of the follo

wing equations represents this situation, where s represents the number of sofas Mr. Spencer sold and e represents the amount he earned?
e = 45s+100
e = 25s+100
e = 45s
e = 35s

2 answers:

Tatiana [17]2 years ago

8 0

First divide the commission by the number sold:

225/5 = 45

300/8 = 37.5 since the commission is a different amount it is not just multiplication so the last two answers are not correct.

Since for 5 the commission was 45 but for 8 it was less than 45, the equation can’t use 45 as a multiplier

The answer has to be E= 25s + 100

stepladder [879]2 years ago

3 0

E=25s+100 that is the answer buddying

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a) P=0.226

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c) P=0.0008

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Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

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b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

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Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

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Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

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We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

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Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

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Read 2 more answers

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Answer:

There are 4,148,350,734,528 ways

Step-by-step explanation:

We have

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56 senators which are Repudiators.

(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?

We want to choose 5 Demonstrators and 5 Repudiators. The number of ways to do this is ${44} \choose {5}$ and $56 \choose 5$ respectively. Therefore, the number of ways to select the committee is given by:

${{44}\choose {5}} \times {{56}\choose{5}}=\frac{44!}{39!5!}\times\frac{56!}{51!5!}=\frac{44!56!}{51!39!5!5!}=\frac{44\times43\times42\times41\times40\times56\times55\times54\times53\times52}{5!5!}=\\\\=\frac{44\times43\times42\times41\times8\times56\times11\times54\times53\times52}{4!4!}= \frac{11\times43\times42\times41\times2\times56\times11\times54\times53\times52}{3!3!}=\\\\\frac{11\times43\times14\times41\times2\times56\times11\times18\times53\times52}{2!2!}=$

$11\times43\times14\times41\times28\times11\times18\times53\times52=4,148,350,734,528$

(b) Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?

<u>If the speaker and vice speaker are chosen between all senators:</u> In this case, the answer will be

$44\times43\times56\times55=5,827,360.$

This is because there are (in the case of Demonstrators) 44 possibilities to choose an speaker and after choosing one, there would be 43 possibilities to choose a vice speaker. The same situation happens in the case of Repudiators.

<u>If the speaker and vice speaker are chosen between the committee:</u> In this case, the answer will be

$5\times4\times5\times4=400$ .

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2 years ago