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3 years ago

Can someone help me with this. Will Mark brainliest. Need answer and explanation/work. Thank you.

Mathematics

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

$14/50\\$

Step-by-step explanation:

The equation for sine is sine = opposite/ hypotenuse

The opposite of W is 14 and the hypotenuse the the side across the 90° angle, which is 50.

So, when you set up the equation it should be $14/50$ .

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All of them I need help please

NemiM [27]

19.

2568.75 - 2500 = 68.75;

( ? / 100 ) * 2500 = 68.75 ;

? * 25 = 68.75;

? = 68.75 ÷ 25

? = 2.75;

r = 2.75 %

7 0

3 years ago

Can somebody tell me how too simplify this expression 3(5*+2)

Pie

Answer:

15+6

Step-by-step explanation:

You would need to multiply 3*5 then 3*2 and then put 15+6

6 0

3 years ago

Read 2 more answers

Use the figure below to find the value of x and PQ.

Scilla [17]

Answer:

x=14

PQ=46

Step-by-step explanation:

PT and TQ are equal, so we can set up the equation like this: PT=TQ

23=2x-5

calculate to find x

x=14

PQ is just PT+TQ so 23+23=46

7 0

3 years ago

Every prime number greater than 10 has a digit in the ones place that is include in wich set of numbers A- 1,3,7,9 B-1,3,5,9 D-1

marta [7]

That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:

-- 0:
   A number greater than 10 with a 0 in the units place is a multiple of
   either 5 or 10, so it's not a prime number.

-- 1:
    A number greater than 10 with a 1 in the units place could be
    a prime (11, 31 etc.) but it doesn't have to be (21, 51).

-- 2:
   A number greater than 10 with a 2 in the units place has 2 as a factor
   (it's an even number), so it's not a prime number.

-- 3:
   A number greater than 10 with a 3 in the units place could be
   a prime (13, 23 etc.) but it doesn't have to be (33, 63) .

-- 4:
   A number greater than 10 with a 4 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 5:
   A number greater than 10 with a 5 in the units place is a multiple
   of either 5 or 10, so it's not a prime number.

-- 6:
   A number greater than 10 with a 6 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 7:
   A number greater than 10 with a 7 in the units place could be
   a prime (17, 37 etc.) but it doesn't have to be (27, 57) .

-- 8:
   A number greater than 10 with a 8 in the units place is an even
   number, and has 2 as a factor, so it's not a prime number.

-- 9:
   A number greater than 10 with a 9 in the units place could be
   a prime (19, 29 etc.) but it doesn't have to be (39, 69) .

So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.

It CAN't have a 0, 2, 4, 5, 6, or 8 .

The only choice that includes all of the possibilities is 'A' .

4 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