Find the value of a. The dot represents the center of the circle

Three friends decide to evenly split their bill at a restaurant. The total bill is $84.99 and they will have to pay an additiona

xz_007 [3.2K]

Given : The Total Bill at the Restaurant is $84.99

Given : The Additional Tax is $5.10

⇒ Total Bill Including the Tax is : (84.99 + 5.10) = $90.09

Given : Three Friends decide to evenly split their Bill

⇒ $90.09 should be Divided into 3 Equal Parts

$\mathsf{\implies The\;Amount\;each\;person\;should\;owe\;is = (\frac{90.09}{3})}$

$\mathsf{\implies The\;Amount\;each\;person\;should\;owe\;is = 30.03\;Dollars}$

His Friend Calculation is not Reasonable

3 0

3 years ago

Evaluate g/-5, when g=-5

daser333 [38]

Answer:

g/-5= 1

Step-by-step explanation:

g=-5, so -5/-5 is 1.

6 0

3 years ago

Read 2 more answers

-2x(6x^4 -7x^2 +x-5 in standard form pls

LuckyWell [14K]

Answer:

-12x^5 + 14x^3 -2x^2 + 10x

Step-by-step explanation:

-2x(6x^4 -7x^2 +x-5) =

-12x^5 + 14x^3 -2x^2 + 10x

8 0

3 years ago

Carl, Gilda, Conroy, and Kyla are the four candidates in a school election. Carl received 1/5 of the votes, Gilda received 5% of

Juliette [100K]

Firstly, we can convert all of the fractions into percentages. To do this, we need to make the denominator of the fraction 100, and whatever we do to the denominator we must also do to the numerator.

5 x 20 = 100
1 x 20 = 20.

So Carl recieves 20/100 or 20% of the votes.

4 x 25 = 100
1 x 15 = 25

So Conroy receives 25/100 or 25% of the votes.

If we add these together and Gilda's 5%, we get 50%. Since there are 100% votes overall, we need to do 100 - 50 = 50.

Kyla receives 50% of the votes.

8 0

3 years ago

Read 2 more answers

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