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Crank
2 years ago
6

HELPPP WILL GIVE BAINLIEST!!!

Mathematics
1 answer:
vazorg [7]2 years ago
8 0

Using the absolute value function, we have that:

  • The gasket with a weight of 0.53 lb should be thrown out.
  • The inequality is: |x - 0.6| ≤ 0.06.

<h3>What is the absolute value function?</h3>

The absolute function is defined by:

  • |x| = x, x ≥ 0.
  • |x| = -x, x < 0.

It measures the distance of a point x to the origin.

For the batch, the weights are given as follows:

0.58, 0.63, 0.65, 0.53, 0.61.

Hence the mean is:

(0.58 + 0.63 + 0.65 + 0.53 + 0.61)/5 = 0.6.

Weights that are more than 0.06 of the pounds must be thrown out, hence the inequality for the valid weights is given by:

|x - 0.6| ≤ 0.06.

Hence the range of valid weighs is:

  • x - 0.6 ≥ -0.06 -> x ≥ 0.54.
  • x - 0.6 ≤ 0.06 -> x ≤ 0.66.

0.53 is not in the interval, hence the gasket with a weight of 0.53 lb should be thrown out.

More can be learned about the absolute value function at brainly.com/question/24837065

#SPJ1

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Answer:

Descriptive statistics

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If, using her class's average where to infer the population average that is inferential statistic. But that is not the case

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4 years ago
What is the unit rate for 240 miles in 6 hours
Amanda [17]
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Nimfa-mama [501]

Answer:

(a) NULL HYPOTHESIS, H_0 : \mu \leq  3.5%

    ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5%

(b) We conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Step-by-step explanation:

We are given that a random sample of seven Ohio banks is selected.The bad debt ratios for these banks are 7, 4, 6, 7, 5, 4, and 9%.The mean bad debt ratio for all federally insured banks is 3.5%.

We have to test the claim of Federal banking officials that the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

(a) Let, NULL HYPOTHESIS, H_0 : \mu \leq  3.5% {means that the the mean bad debt ratio for Ohio banks is less than or equal to the mean for all federally insured banks}

ALTERNATE HYPOTHESIS, H_1 : \mu > 3.5% {means that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks}

The test statistics that will be used here is One-sample t-test;

                T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where,  \bar X = sample mean debt ratio of Ohio banks = 6%

             s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} } = 1.83%

             n = sample of banks = 7

So, test statistics = \frac{6-3.5}{\frac{1.83}{\sqrt{7} } }  ~ t_6

                             = 3.614

(b) Now, at 1% significance level t table gives critical value of 3.143. Since our test statistics is more than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the the mean bad debt ratio for Ohio banks is higher than the mean for all federally insured banks.

Hence, Federal banking officials claim was correct.

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4 years ago
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Dmitry_Shevchenko [17]

Answer: the answer is 88.2 and your welcome!!

Step-by-step explanation:

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6 0
3 years ago
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algol [13]

Answer:

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