Can someone help me on number 30 please

Jason is painting a large circle on one wall of his new apartment. The largest distance across the circle will be 8 feet. Approx

Hunter-Best [27]

Answer:

a) 50.24 ft²

b) 150.72 ft²

Step-by-step explanation:

Jason is painting a large circle on one wall of his new apartment. The largest distance across the circle will be 8 feet.

The distance across a circle is also called a diameter, Hence,

D = Diameter = 8 ft

The Area of a circle = πr²

r = Diameter/2 = 8ft/2 = 4ft

π = 3.14

Hence,

The Area of the circle = π × (4 ft)²

= 50.24 ft²

a) Approximately the amount of wall the circle will cover is 50.24 ft².

b) How much area will he cover if he painted a circle on three walls?

This is calculated as:

1 wall = 50.24 ft²

3 walls = x

Cross Multiply

x = 3 × 50.24 ft²

x = 150.72 ft²

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

Please see the screenshot.

sesenic [268]

Answer:

the answer is c, (-10,4) and the radius is 10

Step-by-step explanation:

Hope this helps!❆

6 0

3 years ago

I need help again with these Please and thank you 17points

Juli2301 [7.4K]