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Wewaii [24]
3 years ago
15

In a class of 666, there are 444 students who are secretly robots.

Mathematics
1 answer:
sveta [45]3 years ago
3 0

Answer:

1/15

Step-by-step explanation:

2/6 for the first one the first the second is 1/5, so 2/6=1/3, so 1/3 x 1/5 = 1/15 your answer is 1/15

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On the drawing for a house plan, the master bedroom is rectangular and has dimensions of 2.25 inches by 2.5 inches. If the scale
bagirrra123 [75]

Answer:

The answer is 16 metric units by the quarter-mile

Step-by-step explanation:  Hope this help

3 0
3 years ago
An ordinary Fair die is a cube with the numbers 1 through 6 on the sides represented by painted spots imagine that such a die is
Hoochie [10]

Answer:  Probability of,  Event A = 5/18

Event B = 1/2

Step-by-step explanation:

Since, The probability of an event = Number of outcomes of the event/ total number of outcomes.

Since, If a fair dire rolled two times then the total number of outcomes, n(S) = 6 × 6 = 36

And, Event A: The sum is greater than 8.

So, possible outcomes are, ( 3,6), (4,5), (4,6), (5,5), (5,6), (5,4), (6,6), (6,5), (6,4), (6,3)

Thus, n(A) = 10

Since, Probability of event A,

P(A) = n(A)/n(S) = 10/36 = 5/18

Therefore, P(A)= 5/18

Now, Event B: The sum is an even number.

So, possible outcomes are, (1,1),(1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,6), (4,2), (4,4), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)

Thus, n(B)= 18

Since, Probability of event B,

P(B)= n(B)/n(S) = 18/36 = 1/2

Therefore, P(B)= 1/2


7 0
3 years ago
A uniform beam of length L = 7.30m and weight = 4.45x10²N is carried by two ovorkers , Sam and Joe - Determine the force exert e
Mama L [17]

Answer:

Effort and distance = Load  x distance

7.30 x 4.45x10^2N = 3.2485 X 10^3N

We then know we can move 3 points to the right and show in regular notion.

= 3248.5

Divide by 2 = 3248.5/2 = 1624.25 force

Step-by-step explanation:

In the case of a Second Class Lever as attached diagram shows proof to formula below.

Load x distance d1 = Effort x distance (d1 + d2)

The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.

If the wheelbarrow is held steady (i.e. in Equilibrium) then the moment of the effort must be equal to the moment of the load :

Effort x its distance from wheel centre = Load x its distance from the wheel centre.

This general rule is expressed as clockwise moments = anti-clockwise moments (or CM = ACM)

 

This gives a way of calculating how much force a bridge support (or Reaction) has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!

The moment of the load on the beam (F) must be balanced by the moment of the Reaction at the support (R2) :

Therefore F x d = R2 x D

It can be seen that this is so if we imagine taking away the Reaction R2.

The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up.

The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example attached.

We know from our experience that a lighter person will have to sit closer to the end of the see-saw to balance a heavier person - or two people.

So if CM = ACM then F x d = R2 x D

from our kitchen scales example above 2kg x 0.5m = R2 x 1m

so R2 = 1m divided by 2kg x 0.5m

therefore R2 = 1kg - which is what the scales told us (note the units 'm' cancel out to leave 'kg')

 

But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated.

Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams (or bridges) have several loads acting upon them.

In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.e. F1=60kg, F2=40kg, d1=6m and d2=9m

 

CM = ACM

(F1 x d1) + (F2 x d2) = R2 x Length of beam

(60kg x 6m) + (40Kg x 9m) = R2 x 12m

(60kg x 6m) + (40Kg x 9m) / 12m = R2

360kgm + 360km / 12m = R2

720kgm / 12m = R2

60kg = R2 (note the unit 'm' for metres is cancelled out)

So if R2 = 60kg and the total load is 100kg (60kg + 40kg) then R1 = 40kg

4 0
2 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
3 years ago
,,,,,....,,,,,:;,,,,,:(
pshichka [43]

start at -2 go over 2 and up 5

(-2, 5)

8 0
3 years ago
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