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m_a_m_a [10]
2 years ago
7

plot two points to show the approximate positions in the number line where Andrea and Larry should tape their cards

Mathematics
1 answer:
Alex17521 [72]2 years ago
8 0

Hello there. To solve this question, we'll have to remember some properties about finding values of expressions.

We need to plot two points in the line to show the approximate positions where Andrea and Larry should tape their cards.

The cards contains a number for each of them.

For Andrea, the number is:

2\pi

To find where Andrea should tape her card, in this case we need to know an approximation for the number pi.

Using 3.141 as an approximation, we have that she should tape her card at:

2\cdot3.141=6.282\approx6.3

For Larry, the number is:

\sqrt[]{17}

In the case of Larry, we can use another idea to find the approximate value.

Remember that for positive numbers, the root of a number is an injective function. It means that for valeus a and b, with a < b, it is true that:

\sqrt[]{a}

Knowing:

16

Taking the square root on both sides of the inequality, we have:

4

In fact, the approximate value of this square root is 4.12, therefore Larry should tape his card at:

4.1

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The sears tower in chicago is 1450 feet high. a model of the tower is 24 inches tall what is ratio
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A rectangle is fenced off for a robot competition. The dimensions of the competition area are 35 feet by 25 feet. To help track
xz_007 [3.2K]

Answer:

The approximate speed of the robot during the 5 seconds is:

  • <u>6 ft/s</u>

Step-by-step explanation:

To calculate the speed of the robot, you must begin with the positions, the first position (3, 18) and the second position (31, 6), you can see, in the height it moved from 3 to 31, it means 28 feet, in the width it moves from 18 to 6, it means 12 feet, with these data you can construct a triangle where you have the opposite leg and adjacent leg, now you must calculate the hypotenuse, because it is the linear path from the first position to the second position that the robot took, for this, you can use the Pythagoras theorem:

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With the value of the distance traveled, and the time used (5 seconds), we can calculate the speed with the next formula:

  • Speed =\frac{distance}{time}
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5 0
3 years ago
Suppose that we want to generate the outcome of the flip of a fair coin, but that all we have at our disposal is a biased coin w
Goshia [24]

Answer:

Step-by-step explanation:

Given that;

the following procedure for accomplishing our task are:

1. Flip the coin.

2. Flip the coin again.

From here will know that the coin is first flipped twice

3. If both flips land on heads or both land on tails, it implies that we return to step 1 to start again. this makes the flip to be insignificant since both flips land on heads or both land on tails

But if the outcomes of the two flip are different i.e they did not land on both heads or both did not land on tails , then we will consider such an outcome.

Let the probability of head = p

so P(head) = p

the probability of tail be = (1 - p)

This kind of probability follows a conditional distribution and the probability  of getting heads is :

P( \{Tails, Heads\})|\{Tails, Heads,( Heads ,Tails)\})

= \dfrac{P( \{Tails, Heads\})  \cap \{Tails, Heads,( Heads ,Tails)\})}{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) }{  {P( \{Tails, Heads,( Heads ,Tails)\}}}

= \dfrac{P( \{Tails, Heads\}) } {  {P( Tails, Heads) +P( Heads ,Tails)}}

=\dfrac{(1-p)*p}{(1-p)*p+p*(1-p)}

=\dfrac{(1-p)*p}{2(1-p)*p}

=\dfrac{1}{2}

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(b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

For a fair coin (0<p<1) , it's certain that both heads and tails at the end of the flip.

The procedure that is talked about in (b) illustrates that the procedure gives head if and only if the first flip comes out tail with probability 1 - p.

Likewise , the procedure gives tail if and and only if the first flip comes out head with probability of  p.

In essence, NO, procedure (b) does not give a fair coin flip outcome.

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