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Angelina_Jolie [31]
3 years ago
6

For a random sample of 50 measurements on the breaking strength of cotton threads, the mean breaking strength was found to be 21

0 grams and the standard deviation 18 grams.
1. Obtain a confidence interval for the true mean breaking strength of cotton threads of this type with confidence coefficient 0.90.
Mathematics
1 answer:
Vikentia [17]3 years ago
6 0

Answer:

205.81 to 214.19

Step-by-step explanation:

Sample size (n) = 50

Sample mean weight (X) = 210 grams

Sample standard deviation (s) = 18 grams

Confidence coefficient (Z) at 90% = 1.645

The confidence interval for the true mean breaking strength of cotton threads of this type, at a 90% confidence level, is given by:

CI=X \pm Z*\frac{s}{\sqrt n}\\CI=210 \pm 1.645*\frac{18}{\sqrt 50}\\CI = 210 \pm 4.19\\CI =205.81\ to\ 214.19

The confidence interval is 205.81 grams to 214.19 grams.

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Answer:

The second statement is Correct.

Step-by-step explanation:

Consider the provided information.

It is given that Antonio uses a calculator to find 38-\frac{44-16}{4} and gets a result of –10.

38-\frac{44-16}{4}=38-\frac{28}{4}

=38-7

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So, the first statement is incorrect.

2nd statement: Antonio found the answer for 38-44-\frac{16}{4}

38-44-\frac{16}{4}=-6-4

38-44-\frac{16}{4}=-10

The second statement is Correct.

Third statement: Antonio found the answer for \frac{38-44-16}{4}

\frac{38-44-16}{4}=-5.5

So, the third statement is incorrect.

Fourth statement: Antonio found the answer for 38+\frac{44-16}{4}

38+\frac{44-16}{4}=38+7

38+\frac{44-16}{4}=45

So, the fourth statement is incorrect.

Hence, the second statement is Correct.

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Step-by-step explanation:

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Does 10 24 26 make a right triangle
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If 14 g of a radioactive substance are present initially and 3 yr later only 7 g​ remain, how much of the substance will be pres
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Explanation:

First, we need to find the values of the sine and cosine of x knowing the value of tan x and x being in the 3rd quadrant. Since tan x = 5/12, using Pythagorean theorem, we know that

\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}

Note that both sine and cosine are negative because x is in the 3rd quadrant.

Recall the addition identities listed below:

\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta

\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x

\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}

\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta

\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x

\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}

\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}

\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}

\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}

Therefore, the expression reduces to

\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}

\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}

\;\;\;\;\;= \dfrac{49}{26}

5 0
2 years ago
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