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Maksim231197 [3]
3 years ago
7

A random sample produced the pie chart below. The chart gives percentages of students that under reported, accurately reported,

and over reported their heights. If there are 2,500 students at the school, how many of them can you reasonably expect report their height accurately?
A. 500
B. 200
C. 50
D. 20

Mathematics
2 answers:
enyata [817]3 years ago
8 0

Answer:

Option A) 500

Step-by-step explanation:

We are given the following information:

Total number of students = 2500

From pie chart:

\text{Under Reported }= \displaystyle\frac{1}{5}\times 100\% = 20\%\\\text{Accurately Reported }= \displaystyle\frac{1}{5}\times 100\% = 20\%\\\text{Over Reported }= \displaystyle\frac{3}{5}\times 100\% = 60\%

We have to find the number of students whose height were accurately reported =

\text{Number of students whose height were accurately reported }= \\\text{Number of students}\times \text{Percentage of students whose height were accurately reported}\\= 2500\times 20\%\\\\= 2500\times \displaystyle\frac{20}{100} = 2500\times \frac{20}{100} = 500

500 students report their height accurately.

tiny-mole [99]3 years ago
7 0

Answer:

The correct option is A.

Step-by-step explanation:

Total number of students at the school is 2,500.

In the given pi chart Pink, Yellow and Sky Blue color represent the percentage of students that under reported, accurately reported, and over reported their heights respectively.

From the graph we can conclude that the yellow portion is approximately one fifth of whole circle. It means the number of students that accurately reported their height is one fifth of total number of students.

2500\times \frac{1}{5}=500

Therefore the number of students that accurately reported their height is 500. Option A is correct.

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The time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minus.
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Answer:

Step-by-step explanation:

Given that the  time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

P(completing exam before 1 hour)

= P(less than an hour) = P(X<60)

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How do I do this? Please explain in detail.
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22.5/(x-6) + 22.5/(x+6) = 9
multiply by x-6
=> (x-6)22.5/(x-6) + (x-6)22.5/(x+6) = 9(x-6)
=> 22.5 + (x-6)22.5/(x+6) = 9(x-6)
multiply by x+6
=> (x+6)22.5 + (x+6)(x-6)22.5/(x+6) = 9(x-6)(x+6)
=> (x+6)22.5 + (x-6)22.5 = 9(x-6)(x+6)
distribute
=> 22.5x+6(22.5) + 22.5x - 6(22.5) = 9(x^2 - 36)
=> 45x = 9x^2 - 9(36)
=> 0 = 9x^2 - 45x - 9(36)
divide by 9
=> 0 = x^2 - 5x - 36
=> 0 = x^2 - 5x - 36
=> 0 = (x - 9)(x + 4)
x=9 and -4
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