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trapecia [35]
2 years ago
9

The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches. Use the Empiric

al Rule to determine the interval and contains the middle 68% of the heights.
a) [40,70]

b)[45,70]

c)[50,60]

d)[45,65]

e)[47,63]

d)none of the above
Mathematics
1 answer:
maks197457 [2]2 years ago
4 0

Answer:  c)[50,60]

Step-by-step explanation:

The Empirical rule says that , About 68% of the population lies with the one standard deviation from the mean (For normally distribution).

We are given that , The heights of students in a class are normally distributed with mean 55 inches and standard deviation 5 inches.

Then by Empirical rule, about 68% of the heights of students lies between one standard deviation from mean.

i.e. about 68% of the heights of students lies between \text{Mean}\pm\text{Standard deviation}

i.e. about 68% of the heights of students lies between 55\pm5

Here, 55\pm5=(55-5, 55+5)=(50,60)

i.e.  The required interval that contains the middle 68% of the heights. = [50,60]

Hence, the correct answer is c) (50,60)

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A treasure map says that a treasure is buried so that it partitions the distance between a rock and a tree in a 5:9 ratio. Marin
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Answer:

A)  (7.6, 8.8)

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Missing Problem Statement:

Given Options:

A) (11.4, 14.2)

B) (7.6, 8.8)

C) (5.7, 7.5)

D) (10.2, 12.6)

I have added the picture showing the traced map onto a coordinate plane to find exact location of treasure.

The coordinates of Tree are (16,21)

Coordinates of rock are (3,2)

Step-by-step explanation:

Let,

Coordinates of treasure be (a,b)

d_{1}= distance from tree to treasure

d_{2}=distance from rock to treasure

d_{1}=\sqrt{(16-x)^2 + (21-y)^2}

d_{2}=\sqrt{x\2 + (y-2)^2}

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Now we just need to cross check by putting the coordinates given in the options one by one to find out value of d_{1},d_{2} and checking if it satisfies the Equation 1.

Check (A) (11.4, 14.2)

d_{2}= 14.8, d_{1}= 8.2,

\frac{d_{2}}{ d_{1}}= 1.8

Check (C) (5.7, 7.5)

d_{2}= 6.13, d_{1}= 16.98,

\frac{d_{2}}{ d_{1}}= 0.36

Check (D) (10.2, 12.6)

d_{2}= 12.8, d_{1}= 10.2,

\frac{d_{2}}{ d_{1}}= 1.25

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d_{2}= 8.2, d_{1}= 14.8,

\frac{d_{2}}{ d_{1}}= 0.55

Which satisfies Equation 1, such that ratio between rock and tree is 5:9 or  \frac{d_{2}}{ d_{1}}=\frac{5}{9}= 0.55

So, the coordinates of the treasure are (B) (7.6, 8.8)

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