Establece el producto cartesiano;representa en diagrama sagital y plano cartesiano M={a,b,c} N={4,8,12} ayuda

2 whole 2/3 times 3/11

sashaice [31]

2 2/3 x 3/11
= 8/4 x 3/11
= 24/44
Simplified
6/11

3 0

3 years ago

Twice a number increased by 2 is zero

Dmitry_Shevchenko [17]

Alright, so 2x+2=0 (with x being the number) is what I'm getting from this. Subtract 2 from both sides to get 2x=-2, and then divide by 2 to get x=-1

7 0

3 years ago

Read 2 more answers

What between 250 and 300

Fiesta28 [93]

Answer:

5 and 0 between 250 and 300

5 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

A = πr2S 360 i need the answer asap solve for S

Tresset [83]

$\sf a=\pi r^2s\times360\\Divide\ by\ \pi r^2\times360\\ \frac{a}{\pi r^2\times360}= \frac{\pi r^2s\times360}{\pi r^2\times360}\\ \frac{a}{\pi r^2\times360}=s\\\\{\boxed{\sf s=\frac{a}{\pi r^2\times360}}$

6 0

3 years ago

Establece el producto cartesiano;representa en diagrama sagital y plano cartesiano M={a,b,c} N={4,8,12} ayuda​

Establece el producto cartesiano;representa en diagrama sagital y plano cartesiano M={a,b,c} N={4,8,12} ayuda