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vladimir1956 [14]
3 years ago
8

2. (15 points) Shelly has a photo that is 7 1/3 inches tall. She wants to shrink it down to fit in a picture frame that is only

5 1/3 inches tall. The photo shop can only reduce photos by certain fractions. They can reduce it to 2/3 the original size, 3/4 the original size, or 5/9 the original size.
A. Which reduction should she use so the picture fills as much of the frame as possible, without being too large?

B. How much extra space is there in the frame when she uses the reduction from Part A?

C. If she had a machine that could reduce by any amount, so that she could make the reduced picture fit in the frame exactly, what fraction would the reduction be?
Mathematics
1 answer:
Inessa [10]3 years ago
5 0

A. Which reduction should she use so the picture fills as much of the frame as possible, without being too large?

Find the scale factor to get rom 7 1/3 inches to 5 1/3 inches:

5 1/3  / 7 1/3 = 0.7272

Now rewrite the fraction as decimals:

2/3 = 0.667

¾ = 0.75

5/9 = 0.555

The closest scale that would still fit the frame would be 2/3 because it is under 0.727.

B. How much extra space is there in the frame when she uses the reduction from Part A?

Multiply the original size by the scale factor to use:

7 1/3 x 2/3 = 4 8/9

Now subtract the scaled size from the original size:

7 1/3 – 4 8/9 = 2 4/9 inches extra

C. If she had a machine that could reduce by any amount, so that she could make the reduced picture fit in the frame exactly, what fraction would the reduction be?

Convert the scale from part A to a fraction:  

0.72 = 72/99 which reduces to 8/11

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Determine whether the improper integral converges or diverges, and find the value of each that converges.
Brrunno [24]

Answer:

Converges at -1

Step-by-step explanation:

The integral converges if the limit exists, if the limit does not exist or if the limit is infinity it diverges.

We will make use of integral by parts to determine:

let:

u=x             dv=e^(2x)\cdot{dx}

du=dx         v=2\cdot{e^(2x)}

\int\limits^a_b {u} \, dv = uv -\int\limits^a_b {v} \, du

\int\limits^a_b {x\cdot{e^2^x} \, dx =2xe^2^x- \int\limits^a_b {2e^2^x} \, dx

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5 0
3 years ago
The growth of one child is represented by the expression 5x – 1. The growth of a second child is represented by the expression 2
IRISSAK [1]

5x-1=2x+3

subtract 2x from both sides.

3x-1=3

add 1 to both sides.

3x=4

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3 years ago
Michelin Tire Company wishes to set a minimum mileage guarantee on its new MX100 tire. Tests reveal the mean mileage is 67,900 w
Keith_Richards [23]

Answer:

The minimum guaranteed mileage is  x = 71489.55

Step-by-step explanation:

From the question we are told that

   The mean is  \mu  =  67900

     The standard deviation is  \sigma =  2050

Generally for the probability to be 4% the minimum guaranteed mileage is evaluated as

    P( X  \ge x ) =1-  P( \frac{X - \mu }{\sigma } < \frac{ x -  67900 }{ 2050}  ) = 0.04

\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )

     P( X  \ge x ) =1-  P( Z

=>  P( X  \ge x ) = P( Z

=>    z = \frac{ x -  67900 }{ 2050}

From the z table, the critical value corresponding to 0.96 to the left of the curve is  

         z = 1.751

So

        1.751 = \frac{ x -  67900 }{ 2050}

=>     x = 71489.55

4 0
3 years ago
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