The figure is drawn on centimeter grid paper. Find the perimeter of the shaded figure to the nearest tenth.
2 answers:
Answer:
Option A. 17.6 cm
Step-by-step explanation:
Now if we redraw the figure as figure attached.
Rectangle ABCD can be divided in three triangles
Now we apply Pythagoras theorem in ΔAED to get the measure of side DE
DE = √(7² + 3²) = √(49+9) = √58 = 7.62 cm
Now we apply Pythagoras theorem in ΔCDF to get the measure of side DF
DF = √(4²+ 3²) = √(16+9) = 5 cm
Now we calculate the perimeter of EBFD
Perimeter of the shaded figure = mEB + mBF + mFD + mDE
= 1 + 4 + 5 + 7.6 = 17.6 cm
Therefore Option A. 17.6 is the answer.
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Answer: p = 0.9337
Step-by-step explanation: from the question, we have that
total number of pen (n)= 15
number of pen that has never been used=9
number of pen that has been used = 15 - 9 =6
number of pen choosing on monday = 3
total number of pen choosing on tuesday=3
note that the total number of pen is constant (15) since he returned the pen back .
probability of picking a pen that has never been used on tuesday = 9/15 = 3/5
probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5
probability of picking a pen that has been used on tuesday = 6/15 = 2/5
probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5
on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .
we know that
probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.
to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution
p(x=r) =
n= total number of pens=15
r = number of pen chosen on tuesday = 3
p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5
q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5
by slotting in the parameters, we have that
p(x=3) =
p(x=3) = 455 *
p(x=3) = 455 * 0.064 * 0.002176
p(x=3) = 0.0633
thus probability that 3 of the pens has been used on tuesday. = 0.0633
probability of ball that none of the 3 pen has never being used on tuesday = 1 - 0.0633 = 0.9337