Mathematics: Please solve all parts of the question, A B C and D

Step-by-step explanation: a) Here 2y + 5 = 21 2y = 21 - 5 = 16 2y = 16y = 16/2 = 8b) angle opposite...

Please solve all parts of the question, A B, №18041910, 17.02.2023 09:34

Mathematics

Richard had just been given a 6-question multiple-choice quiz in his history class. Each question has four answer, of which only one is correct. Since Richard had not attend not attended class recently, he doesn t any of the answer. Assuring that Richard guesses on all six questions, find the indicated probabilities. a. What is the probability that he will answer all questions correctly? b. What is the probability that he will answer at least one questions correctly? c. What is the probability that he will answer at least half questions correctly?

Step-by-step answer

P Answered by PhD

a) 0.02% probability that he will answer all questions correctly.

b) 82.20% probability that he will answer at least one questions correctly

c) 16.94% probability that he will answer at least half questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either Richard answer it correctly, or he answers it wrong. The probability of answering a question correctly is independent from other questions. So we use the binomial probability distribution to solve this problem.

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.

Richard had just been given a 6-question multiple-choice quiz in his history class.

This means that n = 6

Each question has four answer, of which only one is correct. Since Richard had not attend not attended class recently, he doesn't any of the answer.

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

a. What is the probability that he will answer all questions correctly?

This is P(X = 6) .

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

$P(X = 6) = C_{6,6}.(0.25)^{6}.(0.75)^{0} = 0.0002$

0.02% probability that he will answer all questions correctly.

b. What is the probability that he will answer at least one questions correctly?

Either he does not answer any of the questions correctly, or he does answer at least one correctly. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{6,0}.(0.25)^{0}.(0.75)^{6} = 0.1780$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1780 = 0.8220$

82.20% probability that he will answer at least one questions correctly

c. What is the probability that he will answer at least half questions correctly?

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

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

$P(X = 3) = C_{6,3}.(0.25)^{3}.(0.75)^{3} = 0.1318$

$P(X = 4) = C_{6,4}.(0.25)^{4}.(0.75)^{2} = 0.0330$

$P(X = 5) = C_{6,5}.(0.25)^{5}.(0.75)^{1} = 0.0044$

$P(X = 6) = C_{6,6}.(0.25)^{6}.(0.75)^{0} = 0.0002$

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.1318 + 0.0330 + 0.0044 + 0.0002 = 0.1694$

16.94% probability that he will answer at least half questions correctly

StudenGPT

Question 12 (Mandatory) (5 points) The long-run average total cost curve is considered to be an envelope curve as each short-run average total cost curve touches it at only one level of output. Question 12 options: True False Question 13 (Mandatory) (5 points) If the average cost of producing 9 sweaters is $6.50 and the marginal cost of producing the tenth sweater is $6.75, the average cost of producing 10 sweaters will: Question 13 options: A. be $6.50. B. be more than $6.50. C. be less than $6.50. D. be exactly $6.75. Question 14 (Mandatory)

Step-by-step answer

P Answered by Studen AI

1

Question 12: True
Question 13: B
Question 14: B
Question 15: True
Question 16: True
Question 17: True
Question 18: False
Question 19: False
Question 20: C

Mathematics

Here is a simple probability model for multiple-choice tests. Suppose that each student has probability p of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher p than a weak student.) The correctness of an answer to a question is independent of the correctness of answers to other questions. Jodi is a good student for whom p = 0.81. Required: a. Use the Normal approximation to find the probability that Jodi scores 77% or lower on a 100-question test. b. If the test contains 250 questions,

Step-by-step answer

P Answered by Specialist

a) 0.8461 = 84.61% probability that Jodi scores 77% or lower on a 100-question test.

b) 0.9463 = 94.63% probability that Jodi will score 77% or lower.

c) 400 questions.

d) Yes, because the formula is the same, independently of the value of p.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

p = 0.81

Question a:

100 questions means that n = 100

For the approximation, we have that:

$\mu = p = 0.81$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.81*0.19}{100}} = 0.0392$

This probability is the pvalue of Z when X = 0.77. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.77 - 0.81}{0.0392}$

Z = 1.02

Z = 1.02 has a pvalue of 0.8461

0.8461 = 84.61% probability that Jodi scores 77% or lower on a 100-question test.

Question b:

Now n = 250 , so:

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.81*0.19}{250}} = 0.0248$

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.77 - 0.81}{0.0248}$

Z = 1.61

Z = 1.61 has a pvalue of 0.9463

0.9463 = 94.63% probability that Jodi will score 77% or lower.

Question c:

The formula for the standard deviation is:

$s = \sqrt{\frac{p(1-p)}{n}}$

Meaning that it is inversely proportional to the square root of the sample size.

So, to reduce the standard deviation by half, the number of question must be multiplied by (2)^2 = 4.

100*4 = 400

So 400 questions.

d. Laura is a weaker student for whom p = 0.76. Does the answer you gave in (c) for standard deviation of Jodi's score apply to Laura's standard deviation also?

Yes, because the formula is the same, independently of the value of p.

Mathematics

Item 56 Question 1 You answer 24 questions on a 100-point test correctly and earn a 96%. a. All of the questions are worth the same number of points. How many questions are on the test? How many points is each question worth? There are questions on the test. Each question is worth points. Question 2 b. Your friend earns a grade of 76% on the same test. How many questions did your friend answer correctly? Your friend answered questions correctly.

Step-by-step answer

P Answered by Specialist

Question 1:

There are 25 questions on the test.

Each question is worth 4 marks each.

Question 2:

Your friend answered 19 questions correctly

Step-by-step explanation:

Question 1:

question answered = 24

percent earned = 96%

total number of points = 100%

$\frac{questions answered }{percentage earned}$ x 100

$\frac{24}{96}$ x 100 ⇒ 0.25 x 100 ⇒ 25

there are 25 questions on the test

$\frac{100}{25} = 4$

each question is 4 marks each

Question 2:

percent earned by your friend = 76%

(percent earned by your friend x total number of question) ÷ 100

76 x 25 ÷ 100

1900 ÷ 100

= 19

Mathematics

You randomly guess the answers to two questions on a multiple-choice test. Each question has three choices: A, B, and C. a. What is the probability that you guess the correct answers to both questions? $$ Question 2 Write the probability as a percent. Round your answer to the nearest tenth. The probability is about %. Question 3 b. Suppose you can eliminate one of the choices for each question. How does this change the probability that your guesses are correct? It increases the probability that your guesses are correct to 14, or 25%, because you

Step-by-step answer

P Answered by PhD

3

Hello,

This question I've just been jotting down, so I won't explain much. If you want more, just comment!

:D

a. There is only one correct answer for both. Since each question has 3 outcomes, we will have 1 over 3(3). That's 1/9

Thus, a is 1/9.

(about 11.11%

b. If you eliminate one of the choices, then there will be two possible outcomes. We will have 1 over 2(2), so 1/4.

Thus, b is 1/4.

(a in options)

Hope this helps!

(I'm not sure if you just want these answered or the others.)

History

This question is part of a series of two related questions. part a how was government structured in our country’s first constitution? group of answer choices (a) a weak national government and weak state governments (b) a powerful national government and weak state governments (c) a weak national government and powerful state governments (d) a powerful national government and powerful state governments question 14 this question is part of a series of two related questions. part b why was the first government of the united states structured as described

Step-by-step answer

P Answered by Specialist

4

The answers are A for Part A and A and B for Part B.

Hope that helped!

Mathematics

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?

Step-by-step answer

P Answered by Specialist

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.

Mathematics

You randomly guess the answers to two questions on a multiple-choice test. Each question has three choices: A, B, and C. a. What is the probability that you guess the correct answers to both questions? $$ Question 2 Write the probability as a percent. Round your answer to the nearest tenth. The probability is about %. Question 3 b. Suppose you can eliminate one of the choices for each question. How does this change the probability that your guesses are correct? It increases the probability that your guesses are correct to 14, or 25%, because you

Step-by-step answer

P Answered by PhD

3

Hello,

This question I've just been jotting down, so I won't explain much. If you want more, just comment!

:D

a. There is only one correct answer for both. Since each question has 3 outcomes, we will have 1 over 3(3). That's 1/9

Thus, a is 1/9.

(about 11.11%

b. If you eliminate one of the choices, then there will be two possible outcomes. We will have 1 over 2(2), so 1/4.

Thus, b is 1/4.

(a in options)

Hope this helps!

(I'm not sure if you just want these answered or the others.)

Mathematics

In a multiple choice quiz, there are 5 questions each with 4 answer choices (a, b, c, and d). Robin has not studied for the quiz at all and decides to randomly guess the answers. A. What is the probability she gets the majority of the questions correct? B. What is the probability the first question she gets right is the 3rd question? C. What is the probability she gets exactly 3 or exactly 4 questions right?

Step-by-step answer

P Answered by Specialist

A. 0.1035

B. 0.1406

C. 0.1025

Step-by-step explanation:

Given that:

the number of sample questions (n) = 5

The probability of choosing the correct choice (p) = 1/4 = 0.25

Suppose X represents the number of question that are guessed correctly.

Then, the required probability that she gets the majority of her question correctly is:

P(X>2) = P(X=3) + P(X =4) + P(X = 5)

$P(X2) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4} + (^{5}C_{5}) \times (0.25)^5 (1-0.25)^{5-5}$

$P(X2) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} +\dfrac{5!}{5!(5-5)!} \times (0.25)^5 (1-0.25)^{0} \Bigg ]$

P(X>2) = [ 0.0879 + 0.0146 + 0.001 ]

P(X>2) = 0.1035

B.

Recall that

n = 5 and p = 0.25

The probability that the first Q. she gets right is the third question can be computed as:

$P(X=x) = 0.25 ( 1- 025) ^{x-1}$

Since, x = 3

$P(X = 3) = 0.25 ( 1- 0.25 ) ^{3-1}$

$P(X =3) = 0.25 (0.75)^{3-1}$

$P(X =3) = 0.25 (0.75)^{2}$

P(X=3) = 0.1406

C.

The probability she gets exactly 3 or exactly 4 questions right is as follows:

P(X. 3 or 4) = P(X =3) + P(X =4)

$P(X=3 \ or \ 4) = [ (^{5}C_{3}) \times (0.25)^3 (1-0.25)^{5-3} + (^{5}C_{4}) \times (0.25)^4 (1-0.25)^{5-4}]$

$P(X=3 \ or \ 4) = \Bigg [ \dfrac{5!}{3!(5-3)!} \times (0.25)^3 (1-0.25)^{2} + \dfrac{5!}{4!(5-4)!} \times (0.25)^4 (1-0.25)^{1} \Bigg ]$

P(X = 3 or 4) = [ 0.0879 + 0.0146 ]

P(X=3 or 4) = 0.1025

Mathematics

question 1 you answer 24 questions on a 100-point test correctly and earn a 96%. a. all of the questions are worth the same number of points. how many questions are on the test? how many points is each question worth? there are questions on the test. each question is worth points. question 2 b. your friend earns a grade of 76% on the same test. how many questions did your friend answer correctly? your friend answered questions correctly. plz

Step-by-step answer

P Answered by PhD

a) There are Total 25 Question on the test.

Each question is worth 4 points.

b) Your friend answered 19 questions correctly.

Step-by-step explanation:

Given:

Question answered =24

Percent earn = 96%

Total point test = 100

We will first solve for part a.

We need to find Total Question on the test and Points worth each question.

Now We know that percent earn is equal to Question answered divided by Total Question on test and ten multiplied by 100.

Framing in equation form we get;

$Percent\ earn = \frac{\textrm{Question Answered}}{\textrm{Total question on test}}\times 100\\ \\\textrm{Total question on test} = \frac{\textrm{Question Answered}}{\textrm{Perecent earned}}\times 100$