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Kitty [74]
3 years ago
7

Describe how multiplying the width by 14 affects the perimeter and the area of the figure.

Mathematics
1 answer:
Basile [38]3 years ago
6 0

Answer:

Step-by-step explanation:

Area of rectangle=width times height

Perimeter of rectangle=width plus width plus height plus height

If you have a rectangle that is 10 units  wide and 5 units tall.  The area would be 10X5=50units squared and the perimeter would be 10+10+5+5=30 units.

If you multiply the width by 14, everything changes.  10X14 is 140 units.  So the width of the rectangle is now 140 units while the height is 5.  The area would now be 140X5=700 Units squared and the perimeter would be 140+140+5+5=290 units.  

Area of triangle= base time height divided by 2

Perimeter of triangle= side plus side plus side

If its a triangle, let's say a right triangle 10 units at the base and 5 units in height.  The area will be 25 Units squared. 10X5=50 divide by 2=25 units squared.  If you multiply the base by 14....140X5=700 divide by 2=350.

I didn't figure out the perimeter of the triangle, because I'm being too lazy to figure out the length of the third side.

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\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

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\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

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value= (1- \frac{1}{k^2}) \times 100 \% =75\%

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\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

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