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Roman55 [17]
2 years ago
8

On an exam you get two points for every question answered correctly, zero points for each question left blank and lose one-half

point for each question answered incorrectly. what is your total score on the exam if you answered 13 questions correctly, leave 7 questions blank and answer 5 questions incorrectly?
Mathematics
1 answer:
Paraphin [41]2 years ago
3 0

Using multiplication, the total score on the exam if you answered 13 questions correctly, leave 7 questions blank and answer 5 questions incorrectly is 39 point.

According to the question,

For each correct question, you get 2 points, and there are 22 correct questions, so multiply 22 by 2

22 * 2 = 44 points.

For each question left blank, you get 0 points, and there are 7 questions left blank, so multiply 7 by 0

7 * 0 = 0 points

For each incorrect question, you get -1 point, and there are 5 incorrect questions, so multiply 5 by -1

5 * (-1) = -5 points.

The total number of score = 22*2 + 7*0 + 5*(-1) = 44 + 0 - 5 = 39 points.

Hence, using multiplication, the total score on the exam if you answered 13 questions correctly, leave 7 questions blank and answer 5 questions incorrectly is 39 point.

Learn more about multiplication here

brainly.com/question/2865130

#SPJ4

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Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, H_0 : p \leq 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, H_1 : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

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<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

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Answer:

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40°/360° = 1/9

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