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maksim [4K]
3 years ago
9

A blueprint of a plot of land has a scale of 1 1/4 cm to 3 1/4 m the length of the land is 20 cm on the blueprint if i represent

s the length of the actual peace of land. What is i
Mathematics
1 answer:
crimeas [40]3 years ago
8 0

Answer:

52metre

Step-by-step explanation:

Let us use proportion in solving this question, the land has a scale of 1 1/4 cm to 3 1/4 m then the proportion is

( 1 1/4)/ (3 1/4 ) =[blueprint of the plot of land/length of the actual peace of land]

But we need to convert those mixed fractions to improper fraction for easy simplification

1 1/4= [(4×1)+4]= 5/2

3 1/4=[(4×3)+1)]=13/4

Then we can proceed with simplification as

(5/2)/(13/4)=[blueprint of the plot of land/length of the actual peace of land]

But we were given length of the actual peace of land as 20 cm, then

(5/2)/(13/4)=[blueprint of the plot of land/20]

If we cross multiply we have

[blueprint of the plot of land] ×13/4 =5/2 ×20

[blueprint of the plot of land] =(5/2 ×20)÷(13/4)

= 50/(13/4)

=(4×13)/50

= 52

Therefore, the length of the actual peace of land is 52metre

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If f(x) = 2x² - 6x + 8, find f(a - 2).
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The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

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