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Ugo [173]
3 years ago
8

1. A student is taking a multiple-choice exam in which each question has four choices. Assume that the student has no knowledge

of the correct answers to any of the questions. She has decided on a strategy
in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball
for each question and replaces the ball in the box.


The marking on the ball will determine her answer
to the question. There are five multiple choice questions on the exam. What is the probability that she will get:

a. Five questions correct?

b. At least four questions correct?

c. No questions correct?

d. No more than two questions correct?
Mathematics
1 answer:
LenKa [72]3 years ago
3 0

Answer:

a) 0.001 = 0.1% probability that she will get five questions correct.

b) 0.0156 = 1.56% probability that she will get at least four questions correct.

c) 0.2373 = 23.73% probability that she will get no questions correct.

d) 0.8965 = 89.65% probability that she will get no more than two questions correct.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either she gets it correct, or she does not. The probability of getting a question correct is independent of any other question, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

There are five multiple choice questions on the exam.

This means that n = 5

She has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box. She randomly selects one ball for each question and replaces the ball in the box.

This means that p = \frac{1}{4} = 0.25

a. Five questions correct?

This is P(X = 5). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001

0.001 = 0.1% probability that she will get five questions correct.

b. At least four questions correct?

This is:

P(X \geq 4) = P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146

P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001

P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0146 + 0.001 = 0.0156

0.0156 = 1.56% probability that she will get at least four questions correct.

c. No questions correct?

This is P(X = 0). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373

0.2373 = 23.73% probability that she will get no questions correct.

d. No more than two questions correct?

This is:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373

P(X = 1) = C_{5,0}.(0.25)^{1}.(0.75)^{4} = 0.3955

P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2373 + 0.3955 + 0.2637 = 0.8965

0.8965 = 89.65% probability that she will get no more than two questions correct.

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A plane flying horizontally at an altitude of 2 mi and a speed of 540 mi/h passes directly over a radar station. Find the rate a
Rasek [7]

Answer:

The plane's distance from the radar station will increase about 8 miles per minute when it is 5 miles away from it.

Step-by-step explanation:

When the plane passes over the radar station, the current distance is the altitude h = 2. Then it moves b horizontally so that the distance to the station is 5. We can form a rectangle triangle using b, h and the hypotenuse 5. Therefore, b should satisfy

h²+b² = 5², since h = 2, h² = 4, as a result

b² = 25-4 = 21, thus

b = √21.

Since it moved √21 mi, then the time passed is √21/540 = 0.008466 hours, which is 0.51 minutes. Note that in 1 minute, the plane makes 540/60 = 9 miles.

The distance between the plane and the radar station after x minutes from the moment that the plane passes over it is given by the function

f(x) = \sqrt{((9x)^2 + 2^2)} = \sqrt{(81x^2+4)}

We have to compute the derivate of f in x = 0.51. The derivate of f is given by

f'(x) = \frac{162x}{2\sqrt{81x^2+4}}

also,

f'(0.51) = \frac{162*0.51}{2\sqrt{81*0.51^2+4}} = 8.2486

The plane's distance from the station will increase about 8 miles per minute.

4 0
3 years ago
2×2×2+10-4÷4+3×3×3 can y'all please help this is due tomorrow
baherus [9]
Then you'll need to get started on it pretty soon.
Just take it slow and easy, and remember your order of operations (or PEMDAS).
First look through it and do any multiplications and divisions that you find.
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3 years ago
What measurement is equal to 24 kilograms
Vladimir79 [104]
52.9109 in pounds if that's what your asking.
3 0
3 years ago
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Anyone know this?? it’s finding the sine of a triangle!!
LenKa [72]

Answer:

I'm not sure what you want me to answer from this, so I solved for every variable:

Angle A: 83°

Side b: 6.29

Side c: 5.8

Step-by-step explanation:

-----Angle A:

Since the sum of the interior angles of a triangle ALWAYS equal 180°, we can solve for angle A as follows:

A+51+46=180\\A+97=180\\A=83

-----Side b:

Here, we use the sin rule for finding sides, since we know all of the angles as well as one side:

\frac{a}{sin(A)} =\frac{b}{sin(B)} \\\frac{8}{sin(83)} =\frac{b}{sin(51)} \\8.06=\frac{b}{0.78} \\6.29=b

-----Side c:

\frac{a}{sin(A)} =\frac{c}{sin(C)} \\\frac{8}{sin(83)}=\frac{c}{sin(46)} \\8.06=\frac{c}{0.72} \\5.8=c

7 0
2 years ago
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frosja888 [35]
The answer to this question is letter C

We can approach this by first finding the rate of feet per second traveled.

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This answer resembles the slope for answer choice C.

To find the distance traveled in 10 seconds we do 10*8= 80 feet.

I hope this helps, please like and brainliest if correct, thank you!

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