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SVETLANKA909090 [29]
3 years ago
13

According to the FBI Uniform Crime Report, the robbery rate in the United States is 202 per 100,000 people. At that rate, how ma

ny robberies would there be in a state the size of Indiana (5.8 million people)
Mathematics
2 answers:
Tpy6a [65]3 years ago
3 0

Answer:

There would be around 1.2 × 10⁴ robberies.

Step-by-step explanation:

In the United States, the robbery rate is 202 robberies per 100,000 people. We can use proportions to estimate how many robberies would there be in a state like Indiana, with a population of 5.8 million people.

5.8 × 10⁶ people × (202 robberies / 100,000 people) = 1.2 × 10⁴ robberies

jeka57 [31]3 years ago
3 0

Answer:

Step-by-step explanation:

According to the FBI Uniform Crime Report, the robbery rate in the United States is 202 per 100,000 people. This means that for every 100,000 people, there would be 202 robbery incidents.

To determine the number of robberies that would occur in a state like Indiana whose size is 5.8 million people, we would determine how many 100000 are there in 5.8 million and multiply by 202. It becomes

5800000÷100000 × 202 = 58×202 = 11716 robberies

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The width of a rectangle is 5 meters less than its​ length, and the perimeter is 34 meters. Find the length and width of the rec
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Answer:

Length = 11

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Step-by-step explanation:

We know that the length is x and the width is x - 5 since it is 5 less than the length.

The equation to find the perimeter of a rectangle is 2w + 2l

We will plug in the values and solve

2(x-5) + 2(x) = 34

2x - 10 + 2x = 34

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Since the length is simply x, we know that it is 11. We subtract 5 to find the width.

11 - 5 = 6.

The width is 6.

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Piper saw a large box of ping pong balls with 48 balls in each layer
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Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.
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Answer:

a.

T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

b.

T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

Step-by-step explanation:

Remember that for any curve      r(t)  

The tangent vector is given by

T(t) = \frac{r'(t) }{| r'(t)| }

And the normal vector is given by

N(t) = \frac{T'(t)}{|T'(t)|}

a.

For this case, using the chain rule

r'(t) = (  -10*2tsin(t^2) ,   102t cos(t^2)   )\\

And also remember that

|r'(t)| = \sqrt{(-10*2tsin(t^2))^2  +  ( 10*2t cos(t^2) )^2} \\\\       = \sqrt{400 t^2*(  sin(t^2)^2  +  cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t

Therefore

T(t) = r'(t) / |r'(t) | =  (  -10*2tsin(t^2) ,   10*2t cos(t^2)   )/ 20t\\\\ = (  -10*2tsin(t^2)/ 20t ,   10*2t cos(t^2) / 20t  )\\= ( -sin(t^2), cos(t^2) )

Similarly, using the quotient rule and the chain rule

T'(t) = ( -2t cos(t^2) , -2t sin(t^2))

And also

|T'(t)| = \sqrt{  ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t

Therefore

N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

Notice that

1.   |N(t)| = |T(t) | = \sqrt{ cos(t^2)^2  + sin(t^2)^2 } = \sqrt{1} =  1

2.   N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0

b.

Simlarly

r'(t) = (2t,-6,0) \\

and

|r'(t)| = \sqrt{(2t)^2   + 6^2} = \sqrt{4t^2   + 36}

Therefore

T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

Then

T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)

and also

|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 +   ( (3 t)/(9 + t^2)^{3/2})^2  +  0^2 }\\= 3/(t^2 + 9 )

And since

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

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Answer:

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answer = 134,000 / 140  =  957
8 0
3 years ago
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