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Paraphin [41]
4 years ago
7

The scale on a map shows that 10 kilometers is represented by 4 centimeters. Which proportion can be used to find the actual dis

tance, x, represented by 20 centimeters on the map?
Mathematics
2 answers:
earnstyle [38]4 years ago
7 0

Answer: The required proportion is

\dfrac{10}{4}=\dfrac{x}{20}.

Step-by-step explanation:  Given that the scale on a map shows that 10 kilometers is represented by 4 centimeters.

We are to find the proportion can be used to find the actual distance, x, represented by 20 centimeters on the map.

We have

4 centimeters on the map = 10 kilometers.

So, 1 centimeter on the map =\dfrac{10}{4}~\textup{kilometers}.

Also,

20 centimeters on the map = x kilometers.

So, 1 centimeter on the map \dfrac{x}{20}~~\textup{kilometers}.

Therefore, we must have

\dfrac{10}{4}=\dfrac{x}{20}.

Thus, the required proportion is

\dfrac{10}{4}=\dfrac{x}{20}.

Illusion [34]4 years ago
3 0
10/4=x/20
Your answer will be x=50 kilometers
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3 years ago
Assume that the weights of individuals are independent and normally distributed with a mean of 160 pounds and a standard deviati
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Answer:

a) \bf 0.3446^{25}=2.7095*10^{-12}

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Step-by-step explanation:

a)

Since 4300/25 = 172, in average every single person should weight 172 pounds to exceed the design limit.

The probability that one person weights 172 pounds or more is the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of 172.

<em>In Excel and OpenOffice Calc, this value is found with the formula </em>

<em>=1-NORMDIST(172;160;30;1) </em>

<em> (NORMDIST(172;160;30;1) gives the area to the left of 172, so 1-NORMDIST(172;160;30;1) gives the area to the right of 172) </em>

and equals 0.3446

(see picture 1)

Hence, the probability that all the 25 people exceeds 172 pounds equals

\bf 0.3446^{25}=2.7095*10^{-12}

b)

Similarly, we must find a weight w such that if p is the area to the left of w, then  

\bf p^{25}>0.0001  

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\bf p=\sqrt[25]{0.0001}=0.6918

w would be the point such that the area under the Normal curve with mean 160 pounds and standard deviation 30 pounds to the right of w equals 0.6918.

<em>In Excel and OpenOffice Calc this is found with </em>

<em>=NORMINV(1-0.6918;160;30) </em>

and equals 144.97 pounds

(See picture 2)

and the design limit that is exceeded by 25 occupants with probability 0.0001 is 25*144.97 = 3,624.25 pounds

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