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jekas [21]
3 years ago
8

What’s the reason for the fourth statement in this proof

Mathematics
1 answer:
Ksivusya [100]3 years ago
8 0
What statement.........
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What does it mean for data on a scatter plot to have no association? How does that affect the line of best fit?
lilavasa [31]
Извините  можно по подробнее<span />
3 0
3 years ago
TVS are on sale at best buy you decide to buy 3 tvs over the phone and that the total will be $2697.00 but arent sure of each tv
Neko [114]

Answer:

the answer is 899

Step-by-step explanation:

2697 / 3 = 899

899 x 3 = 2697


8 0
3 years ago
  Two​ trains, Train A and Train​ B, weigh a total of 207 tons. Train A is heavier than Train B. The difference of their weights
-BARSIC- [3]

Answer:

A) 184.5 B)22.5

Step-by-step explanation:

207/2= 103.5

A is heavierand the diference of their weights is 81 tons A is 103.5 plus 81 tons or 184.5

And B is 103.5 minus 81 tons or 22.5 tons

5 0
3 years ago
50 POINTS! Reflect quadrilateral PQRS about the y-axis. Then translate the resulting quadrilateral 2 units down.
timama [110]

Answer:

  D.)  P’(12, 6), Q’(8, 7), R’(4, 4), and S’(7, 1)

Step-by-step explanation:

Reflection over the y-axis changes the sign of the x-coordinate:

  (x, y) ⇒ (-x, y) . . . . . reflection over the y-axis

Translation 2 units down, subtracts 2 from the y-coordinate:

  (x, y) ⇒ (x, y -2) . . . . . translation 2 units down

__

The combined transformation is ...

  (x, y) ⇒ (-x, y -2) . . . . . reflection over y-axis and translation 2 down

For point P, the image coordinates are ...

  P(-12, 8) ⇒ P'(-(-12), 8 -2) = P'(12, 6) . . . . matches choice D

5 0
3 years ago
A queue with a single server receives 50 requests per second on average. The average time for the server to address a request is
sdas [7]

Answer:

P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}

Step-by-step explanation:

This is a typical example where the Poisson distribution is a good choice to model the situation.

In this case we have an interval of time of 50 milliseconds as average time for the server to address one request and 50 requests per second.  

By cross-multiplying we determine the expected value of requests every 50 milliseconds.  

We know 1 second = 1,000 milliseconds

50 requests __________ 1000 milliseconds

 x requests __________ 50 milliseconds

50/x = 1000/50 ===> x = 2.5  

and the expected value is 2.5 requests per interval of 50 milliseconds.

According to the Poisson distribution, the probability of k events in 50 milliseconds equals

\bf P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}

5 0
3 years ago
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