List and describe three basic Boolean operations commonly used in computer geometric modeling software

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

3 years ago

Assume a function requires 20 lines of machine code and will be called 10 times in the main program. You can choose to implement

Volgvan

Answer:

"Macro Instruction"

Explanation:

A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.

It is a series of commands and actions that can be stored and run whenever you need to perform the task. You can record or build a macro and then run it to automatically repeat that series of steps or actions.

7 0

3 years ago

A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate

olga2289 [7]

Answer:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

4 0

3 years ago

Suppose values is a sorted array of integers. Give pseudocode that describes how a new value can be inserted so that the resulti

Novosadov [1.4K]

Answer:

insert (array[] , value , currentsize , maxsize )

{

if maxsize <=currentsize

{

return -1

}

index = currentsize-1

while (i>=0 && array[index] > value)

{

array[index+1]=array[index]

i=i-1

}

array[i+1]=value

return 0

}

Explanation:

1: Check if array is already full, if it's full then no component may be inserted.

2: if array isn't full:

Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.
Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.

assign new worth to the position that is next to the known position of initial smaller component.

7 0

3 years ago

A certain robot can perform only 4 types of movement. It can move either up or down or left or right. These movements are repres

Olegator [25]

Answer:

def theRoundTrip(movement):

x=0

y=0

for i in movement:

if i not in ["U","L","D","R"]:

print("bad input")

return

if i=="U":

y+=1

if i=="L":

x-=1

if i=="D":

y-=1

if i=="R":

x+=1

return x==0 and y==0

8 0

4 years ago