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3 years ago

I need help ASAP!!!! Pleaseeee

Mathematics

1 answer:

poizon [28]3 years ago

5 0

Answer:

2nd and 3rd ordered pair are on the graph

Step-by-step explanation:

Substitute the x- coordinate into the equation and if the value obtained is equal to the y- coordinate then it lies on the graph.

(1, 625 )

f(1) = - $25^{1+1}$ = - $25^{2}$ = - 625 ≠ 625 ← No

(0, - 25 )

f(0) = - $25^{0+1}$ = - $25^{1}$ = - 25 ← Yes

(- 1, - 1 )

f(- 1) = - $25^{-1+1}$ = - $25^{0}$ = - 1 ← Yes [ using $a^{0}$ = 1 ]

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Factoring trinomials

Tcecarenko [31]

$c^2$ $2c^2 + 13c +21 \\4c^2 + 13(2)c + 42\\(2c + 6)(2c+7)\\(c+3)(2c+7)$

This is a way to factoring trinomials (there exist different equivalent methods).

Multiply the trinomial but the term accompanying $c^2$ . This is the second line. Then, you could take the square of the $4c^2$ , ant try to create a factor () () that will correspond to the expression in the second line. That is, we want $4c^2 + 13(2) c + 42 = (2c + ?) (2c + ?)$

In ? we put the corresponding numbers that, if we multiply them we will obtain 42, and if we add them we will obtain 13. This numbers are 6 and 7. Then, we have $(2c + 6) (2c +7)$

The last step is divide by the number that we multipy in the first step.

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3 years ago

Tom spends 30% of his monthly budget on rent. His total monthly budget is $3,000. How much does Tom spend on rent?

lilavasa [31]

Hey You!

3,000 * 30% = 900

So, Tom spends $900 on his monthly rent.

5 0

3 years ago

Read 2 more answers

PLS HELP I NEED IT FOR MY TEST. I WILL GIVE BRAINLIEST.

CaHeK987 [17]

Wouldn’t the number of bison decrease??

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 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:

 $P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]$

 $P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\ \\ =15\times0.15\times0.1028=0.2312$

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4 years ago

Solve the inequality w+3/2> 6

gulaghasi [49]

Here is a picture of what I did. If it's not clear, just ask.

Download pdf

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4 years ago