Work out p if r =3 and t=4 P=2r+3t

How can you find the area of a shape with fractional or decimal side lengths?

SOVA2 [1]

Answer:

You convert the fractions to a decimal, nut the rest is still the same

Step-by-step explanation:

3 0

3 years ago

Help pleaseeee!!!! Geometry

puteri [66]

Tan x=opposite/adjacent
cot x=adjacent/opposite

so you need to inverse the fraction
tan x=.78 as a fraction is 78/100
-> inversed= 100/78=1.28=cot x

so it is the second option

8 0

3 years ago

What is the mean for the data set?

PIT_PIT [208]

83.3
Just add up all the numbers and divide by 6, as there are six numbers in this set of numbers

7 0

2 years ago

Read 2 more answers

Find the slope intercept form of slope-2 which passes through (-5,1)

dusya [7]

Answer:

$y=2x+11$

Step-by-step explanation:

So we know that the slope is 2 and the lines passes through the point (-5,1).

We can use the point-slope form. The point-slope form is:

$y-y_1=m(x-x_1)$

Where m is the slope and (x₁, y₁) is a point.

So, let's substitute 2 for m and (-5,1) for (x₁, y₁), respectively. Therefore:

$y-(1)=2(x-(-5))$

Simplify:

$y-1=2(x+5)$

Distribute the 2:

$y-1=2x+10$

Add 1 to both sides:

$y=2x+11$

So, the equation of our line is:

$y=2x+11$

6 0

2 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