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12345 [234]
3 years ago
13

g is constructing a triangular sail for a sailboat. Aluminum supports will run along the horizontal base and vertical height of

the sail, respectively. After making some measurements, Calculo determines that the height of the sail cannot exceed 12 feet, or else it will not be able to fit under the bridges in the area. In addition, the total weight of the supports cannot exceed 80 pounds, or else the boat will sink. If aluminum weighs 5 pounds per foot, what is the area of the largest sail Calculo can build? Fully justify your answer, doing each step you learned in the class prep. In particular, you must show that the area you found is indeed the largest possible.
Mathematics
1 answer:
valkas [14]3 years ago
7 0

Answer: A (max) = 1/2 ( 8 * 8 )   = 32 ft²

Step-by-step explanation:

We have two constrains

a) The height of the sail ( support) cannot exceed 12 feet

b) The total weight of the supports cannot exceed 80 pounds

The weight of a foot of aluminum is 5 .

If we call:

x lenght of base support,  y  height of the vertical support  and the sail shape (is a triangle) we have

A = 1/2* ( x + y )

We know that (given a constant perimeter rectangle, the maximun area  is the square

A = x * y                P = 2x +2y         y = (P - 2x ) ÷ 2  

A = x * ( P - 2x ) ÷ 2  ⇒    A = (Px -2x²) ÷ 2

if we get derivative      A´(x) = (P-4x)/2     ⇒  (P-4x)/2   = 0

P = 4x and    x = P/4

Now we can look an square as two straight triangles joined by the diagonal and these two triangles are of area maximun.

Therefore in our case the sail must be of 8 x 8 feet (base and height)

A (max) = 1/2 ( 8 * 8 )   = 32 ft²   and we will keep the condition of weigh

8 + 8 = 16 feet   and the supports weights  16 * 5  = 80 pounds

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There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit
DaniilM [7]

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) = nCr * p^{r} * q^{n-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) = 15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}

p(x=3) = 455 * 0.4^{3} * 0.6^{12}

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

3 0
3 years ago
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