Select the equation that contains the points (-1, 8) and (3, 4)

Molodets [167]

Mode means ‘most often number’ .

The most often numbers are:
20 and 16

Reason:
20 is shown 3 times.
16 is shown 3 times.
19 is shown 2 times.
18 is shown once.

So, the two most occurring/often numbers are 20 and 16.

Give this answer Brainliest!

3 0

1 year ago

In the number 668 the 6 in the hundreds place is _ the value of the 6 in the tens place

mihalych1998 [28]

6 in the hundreds place = 600
6 in the tens place =60

4 0

3 years ago

Find the range of the function: f(x) = x + 3, for x ≠ 1

Tju [1.3M]

Answer:

All Real Numbers except for 4

Step-by-step explanation:

Objective: Specify domain and range of a function.

The equation is a linear equation. A linear equation range is all real numbers. However x cannot equal 1 so let plug in x to find where the y value CANT be.

$f(1) = 1 + 3 = 4$

So the range is

All Real Numbers except for 4.

3 0

2 years ago

After one hour, the hare had finished 2/3 of a 100-yard race. In that same time, the tortoise had finished 42 3/4 yards. How muc

andriy [413]

Answer:

23 11/12 yards or 23.9166 yards

Step-by-step explanation:

2/3 of 100 yard race is equivalent to 66 2/3 yards. Tge difference between the above and 42 3/4 yards covered by tortoise will be

66 2/3- 42 3/4= 23 11/12 yards or 23.91666 yards

5 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