How do I find the exact product of 379 & 8?

Which of the following rational functions is graphed below?

Shalnov [3]

You didn’t put your whole question so I can’t help sorry :(((((((((((

4 0

3 years ago

On paper, make a rectangle that has points A and B as two of its vertices and has a perimeter of 40 units. Draw and label the tw

Westkost [7]

Answer:

$A(x,y),B(x+10,y),C(x+10,y-10) \:and\: D(x,y-10)$

Step-by-step explanation:

If a Rectangle has a perimeter of 40 units, then we can write that formula and find the a formula for the coordinates:

$\left\{\begin{matrix}w+l=20 & \\ w*l=100 & \end{matrix}\right.\Rightarrow w=20-l\Rightarrow (20-l)l=100\Rightarrow -l^2+20l-100=0\Rightarrow l^2-20l+100\Rightarrow (l-10)(l-10)\Rightarrow l=10\therefore l=10\: and\: w=10$

Therefore we can say:

$A(x,y),B(x+10,y),C(x+10,y-10) \:and\: D(x,y-10)$

Technically when $w=l$ in this rectangle is a square, a degenerate square.

8 0

3 years ago

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

3 years ago

El área de un rectángulo se calcula multiplicando la longitud de la base por la longitud de su altura.

ioda

Para obtener el otro número, debemos dividir la base por el área. Haciendo eso obtenemos 8.6 Para verificar, multiplicamos 10.4 y 8.6. 10.4 x 8.6 = 89.44

Entonces la respuesta es B.) 8.6

4 0

3 years ago

An amusement park's owners are considering extending the weeks of the year that it is opened. The owners would like to survey 10

uranmaximum [27]

The best way to randomly choose the 100 families would be to allow a random number generator to come up with 100 families within a 50 radius of the amusement park.

Using this method would ensure that it is more randomised & not limited to people who come at a specific time or are in a specific area as well as it not be affected by subconscious bias of people when selecting people to survey.

6 0

4 years ago

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