Does the point (2, 6) lie on the circle shown? Explain.

Lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. a bank conducts inter

Otrada [13]

Part A:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85

Thus, the probability that the lie detector will conclude that all 15 are telling the truth if all 15 applicants tell the truth is given by:

 $P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\ \\ =1\times0.0874\times1=0.0874$


Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15

Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

$P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\ \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\ \\ =0.9126$

Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

$\mu=npq \\ \\ =15\times0.15\times0.85 \\ \\ =1.9125$

Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The probability that the number of truthful applicants classified as liars is greater than the mean is given by:

 $P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]$

 $P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\ \\ =15\times0.15\times0.1028=0.2312$

8 0

2 years ago

Simplify. (y3)5 · y4

ICE Princess25 [194]

y^3 * 5 * y^4

= 5y^(3+4)

= 5y^7 answer

8 0

2 years ago

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If f(x) = x^2-1 which of the following ordered pairs would be on the graph of f(x)

Alex Ar [27]

These following ordered pairs are on the graph:

(0,1)
(5, 24)
(-2,3)

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An ordered pair (x, f(x)) will be on the graph if replacing them into the equation, we get an equivalence.

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The equation is:

$f(x) = x^2 - 1$

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(0,1)

Here, we test $x = 0, f(x) = 1$

Replacing into the equation:

$1 = 0^2 - 1$

$1 = -1$

False, so not on the graph.

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(0,1)

Here, we test $x = 1, f(x) = 0$

Replacing into the equation:

$0 = 1^2 - 1$

$0 = 0$

True, so on the graph.

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(3,5)

Here, we test $x = 3, f(x) = 5$

Replacing into the equation:

$5 = 3^2 - 1$

$5 = 8$

False, so not on the graph.

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(5,24)

Here, we test $x = 5, f(x) = 24$

Replacing into the equation:

$24 = 5^2 - 1$

$24 = 24$

True, so on the graph.

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(-2,3)

Here, we test $x = -2, f(x) = 3$

Replacing into the equation:

$3 = (-2)^2 - 1$

$3 = 3$

True, so on the graph.

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(-4,-17)

Here, we test $x = -4, f(x) = -17$

Replacing into the equation:

$-17 = (-4)^2 - 1$

$-17 = 15$

False, so not on the graph.

A similar question is given at brainly.com/question/2102317

6 0

2 years ago

On a backpacking trip through Europe, Troy spent 20 Swiss francs to stay one night in a youth hostel. If the foreign exchange ra

weqwewe [10]

Every US Dollar equals $2.2 in France so 9.09.

5 0

2 years ago

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Quadrilateral QRST is similar to quadrilateral EDGF. What's the scale factor from EDGF to QRST

mart [117]

Answer:

C. $\frac{6}{5}$

Step-by-step explanation:

The ratio of the corresponding sides of a polygon is said to be equal. The ratio gives us the scale factor.

Therefore, assuming that quadrilateral QRST was the original polygon that was scaled up to give a similar polygon, EDGF, the scale factor would be expressed as $\frac{42}{35}$

The scale factor = $\frac{42}{35} = \frac{6*7}{5*7} = \frac{6}{5}$

5 0

2 years ago