105 +35x written in another way __ (_

I need help with this question.

pashok25 [27]

Answer:

B. 9/91

Step-by-step explanation:

-81/9 = -2

9/81 = 1/9

√9 = 3

√81 = 9

B is the odd one out; it's a fraction.

6 0

3 years ago

Help me with this math homework

sp2606 [1]

Answer:

1.) 2 over 5

2.)7.5 over 50

3.)1 1/2

5.)0.95%

6.)2.5%

7.)0.94

i cant read 9 and ten

11.)61 over 100

12.)7 over 25

13.)207 over 1000

15.)14 over 25

16.)13 over 50

17.)3 over 500

19.)5 over 8

20.)42 over 125

21.)3 over 250

I tried my best!! hope this helps you out

5 0

4 years ago

Zoe is 16 years old. Her brother, Luke, is 3 years more than half her age. Write a numerical expression for Luke's age.

olga nikolaevna [1]

Answer:

16 : 2 + 3 (remember pemdas)

Step-by-step explanation:

Zoe is 16 years old. Her brother, Luke, is 3 years more than half her age. Write a numerical expression for Luke's age.

16 : 2 + 3 = 11

8 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

4 years ago

Add: (-3p6 + 5) + 9p6

timama [110]

Answer:

6p^6+ 5

Step-by-step explanation:

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((0 - (3 • (p6))) + 5) + 32p6

Step 2 :

Equation at the end of step 2 :

((0 - 3p6) + 5) + 32p6

Step 3 :

Trying to factor as a Sum of Cubes :

3.1 Factoring: 6p6+5

4 0

3 years ago

105 +35x written in another way __ (__)

105 +35x written in another way ()