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UNO [17]
3 years ago
5

How are real numbers used to describe real-world situations?

Mathematics
1 answer:
Alinara [238K]3 years ago
7 0
Real numbers are all the rational numbers such as -7 and 4/3 and all irrational numbers. Real numbers are used mainly in counting and are used in measurement which is always present in real-world situations. These numbers help us to quantify things around.
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Past experience has indicated that the breaking strength of yarn used in manufacturing drapery material is normally distributed
vlabodo [156]

Answer:

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm)

Step-by-step explanation:

Our sample size is 11.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 11-1 = 10.

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 10 and 0.025 in the two-sided t-distribution table, we have T = 1.812

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

s = \frac{4}{\sqrt{11}} = 1.2060

Now, we multiply T and s

M = Ts = 1.812*1.2060 = 2.19/tex]For the lower end of the interval, we subtract the sample mean by M. So the lower end of the interval here is[tex]L = 97 - 2.19 = 94.81 = 94.8cm

For the upper end of the interval, we add the sample mean and M. So the upper end of the interval here is

L = 97 + 2.19 = 99.19 = 99.2cm

So

95% two-sided confidence interval on the true mean breaking strength is (94.8cm, 99.2cm).

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Fittoniya [83]

1/9 is the answer for that question

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A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100.
Crank

Answer: 1.25

Step-by-step explanation:

Given: A college-entrance exam is designed so that scores are normally distributed with a mean(\mu) = 500 and a standard deviation(\sigma) =  100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = \dfrac{Y-\mu}{\sigma}

Z-score = \dfrac{625-500}{100}

⇒ Z-score = \dfrac{125}{100}

⇒Z-score =1.25

Therefore , the required z-score = 1.25

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Answer:

3686

Step-by-step explanation:

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