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Alekssandra [29.7K]
3 years ago
15

subtract 6 from me then multiply by 2 if you subtract 40 and then divide by 4 you get 8 what number am I

Mathematics
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

The answer is 11

Step-by-step explanation:

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d

Step-by-step explanation:

i did this one

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Help me out please .
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I’m pretty sure it’s A but it could also me C
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Due to weather mark walk to school day of the 20 days in february what would be the fraction representation for the number of da
nekit [7.7K]

Tbe question doesn't state the number of days he walked to school out of 20 school days in the

Month of February.

Answer:

Kindly check explanation

Step-by-step explanation:

To. Obtain the fractional representayion for the number of days he walked to school ;

Let the number of days he walked to school = 10 (this is an hypothesized value)

Number of school days in February = 20

Fraction = 10 /20 = 0.5

The percentage of times walked to school : multiply the fraction obtained above by 100%

0.5 * 100% = 50%

Just put the desired value and proceed uSing the steps described above.

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
Please help! 1 question! 50 points! Show your work!
KatRina [158]

Answer:

  \dfrac{2x}{3a^2}

Step-by-step explanation:

\dfrac{3a^4b}{2x}\cdot\dfrac{4x^2}{9a^6b}=\dfrac{3\cdot 4}{2\cdot 9}a^{4-6}b^{1-1}x^{2-1}=\dfrac{2x}{3a^2}

The applicable rules of exponents are ...

  • (a^b)(a^c) = a^(b+c)
  • a^-b = 1/a^b
7 0
3 years ago
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