Which number is divisible by 2, 3, 4, and 6? a.124 b.3,072 c.1,005 d.786

If u answer this I'll make a random question with a 100 points on it so u can get it

mario62 [17]

Answer:

5 and a half hours

Brainliest maybe?

3 0

3 years ago

Read 2 more answers

The ratio of chaperones to students on a field trip to New York City is 2 to 11. If there are 297 students on the trip, how many

Olenka [21]

297/11 = 27

So we have 27 groups of 11 students. If there are 2 chaperones for every 11 students, then we can do:

27x2=54

There are 54 chaperones on the field trip.

7 0

3 years ago

How do I find the measurement of this ARC m<IEF

andrey2020 [161]

The measure of a central angle is equal to measure of a minor arc. That makes m<GEH=17x+12. By the Vertical Angles Theorem, m<GEH and m<IEF are equal to each other (m<GEH=17x+12=m<IEF). By the same theorem, m<FEG and m<IEH are also equal (m<FEG=8x-7=m<IEH). The angles in a circle must all add up to 360 degrees, 2(17x+12)+2(8x-7)=360. Solve for x, then plug x into the equation for m<IEF.
Hope this helps!

4 0

4 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

4 years ago

Four movie tickets cost $36. At this rate, what is the cost of 5 and 11 movie tickets Please answer!

MrRissso [65]

For 5 movie tickets it’s $45 and for 11 it’s $99

6 0

3 years ago