P of selecting point on the shaded region = shaded area/whole area
<span>P( selecting point on the shaded ) = ( the four shaded circles ) / the whole square </span>
<span>P of selecting point on the shaded = ( 4 * ( π * r^2 ) )/ x^2 </span>
<span>P of selecting point on the shaded = ( 4 * ( π * (x/4)^2 ) )/ x^2 </span>
<span>P of selecting point on the shaded = ( 4 * ( π * x^2/16 ) )/ x^2 </span>
<span>P of selecting point on the shaded = ( π * x^2/4 )/ x^2 </span>
<span>P of selecting point on the shaded = x^2( π/4 )/ x^2 </span>
<span>P( selecting point on the shaded ) = π/4 ≈ 0.7854 ≈ 79%
=80%
D is right option hope this helps</span>
The answer is: " 1024π units² " ; or; " 3215.36 units² " .
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A = π * r² ;
A = π * 32² = 1024 π units² ;
or, using "3.14" for "π" ;
1024 * π = 1024 * (3.14) = 3215.36 units² .
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Answer:
$286
Step-by-step explanation:
220:100=2.2
2.2 x 30 = 66
220+66=286
Answer:
length of the longest pencil that can fit must be less than 13cm
Step-by-step explanation:
To get the longest pencil that fit the case, we will use the pythagoras theorem;
diameter = 10cm
radius = 10/2
radius = 5cm
Height = 12cm
Get the length of the longest side
l² = r²+h²
l² = 5²+12²
l² = 25 + 144
l² = 169
l = √169
l = 13cm
Hence the length of the longest pencil that can fit must be less than 13cm
Answer:
its D
Step-by-step explanation:
good luck
