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zysi [14]
3 years ago
11

Rudy worked hard all summer. Before school starts he decided to go shopping and spend some of his earnings. He spent one third o

f his money on a new shirt at Hot Fashions in the mall. He then went to Marvin's Music and bought a CD with one third of the money he had left. At Movies-to-go he spent one third of the money he now had on a video he wanted. He stopped one more time on his way home and bought a paperback at Bert's Book Shop with one third of his remaining money. Rudy arrived home with just $16.00 in his wallet. How much money did he have when he started out?
Mathematics
1 answer:
Rudiy273 years ago
6 0

Answer:

This is a constant problem. He kept on spending 1/3 of his money. The inverse number of 1/3 is 3. Basically the fraction flipped. So he spent 1/3 of his money 4 times. So we multiply 16*3*3*3*3. Multiply by 3 four times. 16*3 is 48. 48*3 is 144. 144*3 is 432. 432*3 is 1296. He started with 1296 dollars. If hes a kid hes rich.

Step-by-step explanation:


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Read 2 more answers
In a class of 19 students, 3 are math majors. A group of four students is chosen at random. (Round your answers to four decimal
KatRina [158]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

nCx=\frac{n!}{x!(n-x)!}

So, the number of ways in which we can select four students from a group of 19 students is:

19C4=\frac{19!}{4!(19-4)!}=3,876

On the other hand, the number of ways in which we can select four students with no math majors is:

(16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

(16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680

(16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360

(16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16

Then, the probability that the group has no math majors is:

P=\frac{1820}{3876} =0.4696

The probability that the group has at least one math major is:

P=\frac{1680+360+16}{3876} =0.5304

The probability that the group has exactly two math majors is:

P=\frac{360}{3876} =0.0929

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3 years ago
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