An icosahedron has twelve faces, the most of any Platonic solid.

What is the solution to the system of equations shown below?

Alexxandr [17]

Answer:

b. (1, 3, -2)

Step-by-step explanation:

A graphing calculator or scientific calculator can solve this system of equations for you, or you can use any of the usual methods: elimination, substitution, matrix methods, Cramer's rule.

It can also work well to try the offered choices in the given equations. Sometimes, it can work best to choose an equation other than the first one for this. The last equation here seems a good one for eliminating bad answers:

a: -1 -5(1) +2(-4) = -14 ≠ -18

b: 1 -5(3) +2(-2) = -18 . . . . potential choice

c: 3 -5(8) +2(1) = -35 ≠ -18

d: 2 -5(-3) +2(0) = 17 ≠ -18

This shows choice B as the only viable option. Further checking can be done to make sure that solution works in the other equations:

2(1) +(3) -3(-2) = 11 . . . . choice B works in equation 1

-(1) +2(3) +4(-2) = -3 . . . choice B works in equation 2

7 0

3 years ago

Please help I really don’t know what to do

just olya [345]

Answer:

The answer is 0

Step-by-step explanation:

2 * 2 * 2 = 8

8 * 0.5 = 4

4 - 4 = 0

0 / 2 = 0

Make sure the 0.5 is actually 0.5, because it is hard to tell in that expression. if anything, anything, the steps are the same and all you would need to do is the change the numbers.

Hope this helps :)

7 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

Kristen is 64 inches tall, and she stands 12 feet away from a streetlight. If she casts an 82-inch-long shadow, how tall is the

o-na [289]

Answer:

176.39 inches or

14.70 feet

Step-by-step explanation:

Consider the right triangle made by Kristen, ground and shadow.

This triangle has one leg as 64 inches.

Next consider the right triangle formed by street light, ground upto shadow tip.

The two triangles have common angle of elevation and also another angle as 90 degrees.

Hence the two triangles would be similar

Also if A is the angle made by hypotenuse of both triangles with the ground we have

$tanx=\frac{64}{82}$

This value also equals by bigger triangle as

$tanx=\frac{h}{82+12(12)}=\frac{h}{226}$

From these two we get

h = height of street light = $\frac{226(64)}{82} =176.39$

6 0

3 years ago

How many answers are in each of these three questions: 1. x squared minus three x plus four equals zero 2. Two x squared plus fo

vaieri [72.5K]

1. No solutions ( no answers) 2 . 2 answers 3. 1 answer

6 0

3 years ago