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In-s [12.5K]
3 years ago
6

Help me with question 4 and 5

Mathematics
1 answer:
lisov135 [29]3 years ago
3 0

Answer:

2.6768·10^(-14)  (Answer C)

Step-by-step explanation:

Problem 4:  We ADD the powers of 10:  10^6·10^(-3) = 10^3.  Also, 3.6·1.7 = 6.12.  Therefore, Answer D is correct.

Problem 5:  Again we ADD (combine) the powers of 10:  10^(-10)·10^(-5) =

10^(-15).  4.78·5.6 = 26.768.  Our intermediary answer is thus:

26.768·10^(-15).  This must be rewritten in the form x.xxxx, so here we have:

2.6768·10^(-14)  (Answer C)

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A triangle has the following vertices (-1,1), (6,-2)and (3,5) if the scale factor of 3, what will be the vertices of the image
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Answer:

  a.  (-3,3), (18,-6), (9,15)

Step-by-step explanation:

If the scale factor is 3 and dilation is about the origin, each of the coordinates of the pre-image is multiplied by 3 to get the coordinates of the image.

Multiplying (-1, 1) by 3 gives (-3, 3), which matches choices A and C.

Multiplying (6, -2) by 3 gives (18, -6) which matches choices A and D.

The only viable choice is choice A.

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Which of the m values satisfy the following the inequality
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Two dice are thrown. Let E be the event that the sum of the dice is
bearhunter [10]

Answer:

  • P(E) = 1/2
  • P(F) = 11/32
  • P(G) = 1/6
  • P(EF) = 5/52
  • P(FG) = 1/32
  • P(EG) = 1/6

Step-by-step explanation:

For the sum to be even, both dice can be odd, or both even. The probability of a dice being odd is 1/2 and the same is for it to be even. Since the result of the dices are independent, we have that

P(E) = (1/2)² + (1/2)² = 1/2

Out of the 36 possible outcomes for the dice (assuming that you can distinguish between first and second dice), there are 11 cases in which one dice is a 6 (if you fix 1 dice as 6, there are 6 possibilities for the other, but you are counting double 6 twice, so you substract one and you get 6+6-1 = 11). Since all configurations for the dices have equal probability, we get that

P(F) = 11/32

The probability for the second dice to be equal to the first one is 1/6 (it has to match the same number the first dice got). Hence

P(G) = 1/6

for EF, you need one six and the other dice even. For each dice fixed as 6 we have 3 possibilities for the other. Removing the repeated double six this gives us 5 possibilities out of 32 total ones, thus

P(EF) = 5/32

If one dice is 6 and both dices are equal, then we have double six, as a result there is only one combination possible out of 32, therefore

P(FG) = 1/32

If both dices are equal, in particular the sum will be even, this means that G= EG, and as a consecuence

P(EG) = P(G) = 1/6

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