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mihalych1998 [28]
3 years ago
8

How many solutions does the equation have? If you answer correctly you get a good amount of points, please give a good detailed

answer to receive all points.
4x+2(x-3) = 8x+12
Mathematics
2 answers:
zmey [24]3 years ago
6 0
There is one solution because x has a power of 1
if u want to solve it:
4x+2x-6= 8x+12
6x-6= 8x+12
2x+12=-6
2x=-18
x=-9
creativ13 [48]3 years ago
4 0
4x + 2x - 6 = 8x + 12 | -8x + 6
-2x = 18 | ÷ (-2)
x = -9
that's it :)
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lora16 [44]

Answer

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Explanation

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Answer:

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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 <span>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 </span>the lie detector will conclude that all 15 are telling the truth if <span>all 15 applicants tell the truth is given by:

</span>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
<span>

</span>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 <span>probability that the number of truthful applicants classified as liars is greater than the mean is given by:

</span>P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\  \\ 1-[P(0)+P(1)]
<span>
</span>P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\  \\ =15\times0.15\times0.1028=0.2312<span>
</span>
8 0
3 years ago
George said if 2 and 4 are factors of a number, then 8 is a factor of the whole number. is he correct explain
Mamont248 [21]
He is correct and incorrect. 2 and 4 can create factors but they can also be factors of other numbers. So 8 can be a factor of bigger whole numbers and can also have it own smaller factors.
8 0
3 years ago
A company wants to determine the level of interest among its employees for an annual summer picnic. The company wants to survey
Alja [10]

Answer: Use employee identification numbers to randomly select 200 employees

Step-by-step explanation:

Random sampling refers to a sampling technique whereby each sample has an equal chance of being selected. It is an unbiased representation of the entire population and this is vital in drawing conclusion.

From the options given, the best way to randomly choose these 200 employees will be to use employee identification numbers to randomly select 200 employees.

7 0
3 years ago
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