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Elden [556K]
3 years ago
11

Shortly after the introduction of a new​ coin, newspapers published articles claiming the coin is biased. The stories were based

on reports that someone had spun the coin 200200 times and gotten 108108 headsminus−​that's 5454​% heads. a )font size decreased by 1a) Use the Normal model to approximate the Binomial to determine the probability of spinning a fair coin 200200 times and getting at least 108108 heads. b )font size decreased by 1b) Do you think this is evidence that spinning this new coin is​ unfair? Would you be willing to use it at the beginning of a sports​ event? Explain.
Engineering
1 answer:
garik1379 [7]3 years ago
3 0

Answer:

(a) 0.12924

(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

Explanation:

(a)

n=200 for fair coin getting head, p= 0.5

Expectation = np =200*0.5=100

Variance = np(1 - p) = 100(1-0.5)=100*0.5=50

Standard deviation, s = \sqrt {variance}=\sqrt {50}= 7.071068

Z value for 108, z =\frac {108-100}{7.071068}= 1.131371

P( x ≥108) = P( z >1.13)= 0.12924

(b)

Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.

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Explanation:

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Now meter stick can read to nearest millimeter.

It is given that length is to be measured with a precision of 1% of 20mm= 1/100 * 20= 0.2mm

Since the least count is 1mm of meter stick and precision required is less than that. So, meter stick cannot be used for this, travelling microscope can be used for this as it can read to 0.1mm.

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Answer:

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The time at which they pass each other is 1.6 sec.

Explanation:

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Using 2nd eqn. of motion:

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(h)a = 7t - 4.9t² + 24  ______ eqn (1)

For ball B, we have:

height = (h)b

Initial Velocity = Vi = 22 m/s

g = - 9.8 m/s² (for upward motion)

Using 2nd eqn. of motion:

h = Vi t + (1/2)gt²

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(h)b = 22t - 4.9t²    ______ eqn (2)

Now, when the too balls pass each other, there height must be same.

Therefore,

(h)a = (h)b

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7t - 4.9t² + 24 = 22t - 4.9t²

22t - 7t = 24

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<u>t = 1.6 sec</u>

Now, for the height, at which they pass each other put t = 6sec in eqn (2)

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h = (22)(1.6) - (4.9)(1.6)²

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<u>h = 22.656 m</u>

<u></u>

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Answer:

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S is the space between vines, in feet.

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grapvines=(rowLength-2*amountSpace)/vinesSpace;

• Finally, display the number of grapevines that will fit in the row.

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