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liq [111]
3 years ago
9

Find Percent of a Number - Instruction - Level F

Mathematics
2 answers:
dimaraw [331]3 years ago
6 0

Answer:

Sierra's team has a greater percent of left-handed batters.

25% of Sierra's team bat left handed and 20% of Aubrey's team bat left handed.

Step-by-step explanation:

Sierra: 7/28 = 1/4

1/4 written as a percentage is 25%

Aubrey: 5/25 = 1/5

1/5 written as a percentage is 20%

25% is greater than 20%, so Sierra's team has a greater percent of left-handed batters.

25% of Sierra's team bat left handed and 20% of Aubrey's team bat left handed.

Hope this helps ;)

Serhud [2]3 years ago
4 0
<h3>Question A: </h3>

Which team has a greater percent of left-handed batters?

<h3>Solution:</h3>

Sierra's team has a greater percentage of left-handed batsmen than Aubrey. Below is the work.

<u>Aubrey's team: 5/25 left-handed batsmen</u>

  • => 5/25 x 100 = 1/5 x 100
  • => 20%

<u>Sierra's team: 7/28 left-handed batsmen</u>

  • => 7/28 x 100
  • => 1/4 x 100
  • => 25%

<u>Hence, Sierra's team has a greater percentage of left-handed batsmen than Aubrey. </u>

<h3>Question B: </h3>

What percent of the players on Sierra's team bat left-handed?

<h3>Solution:</h3>

<u>25% of the players on Sierra's team are left-handed batsmen.</u>

Below is the work.

  • Sierra's team: 7/28 left-handed batsmen
  • => 7/28 x 100
  • => 1/4 x 100
  • => 25%

<u>Hence, 25% of the players on Sierra's team are left-handed batsmen.</u>

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alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

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Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

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x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

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<h3>Given:</h3>

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<h3>To Find:</h3>

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<h3>Solution:</h3>

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