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noname [10]
2 years ago
14

Marcus has three random samples of the same size from a population of 700 people. In one sample, 53% of people have a library ca

rd. In another, 51% of people have a library card. In the third, 43% have a library card. Based on the three samples, how many people with a library card should he expect are in the population?
a.301
b.357
c343
d.371
pls answer quick
Mathematics
1 answer:
Lapatulllka [165]2 years ago
5 0

Answer:

343

Step-by-step explanation:

Finding? How many people with a library card should he expect in the population?

Important? Population = 700. Sample 1= 53%. Sample 2 = 51%. Sample 3 = 43%.

How? Average the 3 samples percentages together. The percent sign tells us what the sample size is (% = out of 100). Set up the proportions for library card to sample size and then library card to population. Cross multiply.

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In your own words, explain why when we multiply terms that have the same bases, we add the exponents.
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2 years ago
The prices of commodities X,Y,Z are respectively x, y, z, rupees per unit. Mr. A purchases 4 units of Z and sells 3 units of X a
liubo4ka [24]

Answer:

(x,y,z)=(1477, 1464, 1437)

Step-by-step explanation:

Consider the selling of the units positive earning and the purchasing of the units negative earning.

<h3>Case-1:</h3>
  • Mr. A purchases 4 units of Z and sells 3 units of X and 5 units of Y
  • Mr.A earns Rs6000

So, the equation would be

3x  +  5y - 4z = 6000

<h3>Case-2:</h3>
  • Mr. B purchases 3 units of Y and sells 2 units of X and 1 units of Z
  • Mr B neither lose nor gain meaning he has made 0₹

hence,

2x   - 3y  +  z = 0

<h3>Case-3:</h3>
  • Mr. C purchases 1 units of X and sells 4 units of Y and 6 units of Z
  • Mr.C earns 13000₹

therefore,

- x    + 4y  +  6z = 13000

Thus our system of equations is

\begin{cases}3x  +  5y - 4z = 6000\\2x   - 3y  +  z = 0\\ - x    + 4y  +  6z = 13000\end{cases}

<u>Solving </u><u>the </u><u>system </u><u>of </u><u>equations</u><u>:</u>

we will consider elimination method to solve the system of equations. To do so ,separate the equation in two parts which yields:

\begin{cases}3x  +  5y - 4z = 6000\\2x   - 3y  +  z = 0\end{cases}\\\begin{cases}2x   - 3y  +  z = 0\\ - x    + 4y  +  6z = 13000\end{cases}

Now solve the equation accordingly:

\implies\begin{cases}11x-7y=6000\\-13x+22y=13000\end{cases}

Solving the equation for x and y yields:

\implies\begin{cases}x= \dfrac{223000}{151}\\\\y= \dfrac{221000}{151}\end{cases}

plug in the value of x and y into 2x - 3y + z = 0 and simplify to get z. hence,

\implies z= \dfrac{217000}{151}

Therefore,the prices of commodities X,Y,Z are respectively approximately 1477, 1464, 1437

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