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AURORKA [14]
3 years ago
10

Find the cost of flooring a square shaped room of side 10m at a rate of Rs.50 per square meter.

Mathematics
1 answer:
Ierofanga [76]3 years ago
6 0
<h3>Explanation:</h3>

We have to find the area of the square shaped room.

<h3>Given:</h3>

Side of the square shaped room = 10 m

The cost of flooring a square metre is Rs. 50

<h3>To Find:</h3>

The cost of flooring a square shaped room.

<h3>Formula:</h3>

Area of square = Side × Side

<h3>Solution:</h3>

Area of the square = 10 × 10 = 100 sq.m

1 sq.m = Rs. 50

100 sq.m = 50(100) = Rs. 5,000

<h2>Answer:</h2>

The cost of flooring a square shaped room of side 10 m is <u>Rs. 5,000</u><u>.</u>

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Answer:

a) P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

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And using the z score we got:

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c) P(X

And if we use the continuity correction we got:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(25,5)  

Where \mu=25 and \sigma=5

Part a

For this case we want to find this probability:

P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)

And if we use the z score given by:

z =\frac{x-\mu}{\sigma}

We got this:

P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)

And we can find this probability like this:

P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728

Part b

For this case we want this probability:

P(X \leq 30)

And if we use the continuity correction we got:

P(X \leq 30)= P(X

And using the z score we got:

P(X

Part c

For this case we want this probability:

P(X

And if we use the continuity correction we got:

P(X

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