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Veronika [31]
3 years ago
13

What is the answer to 12 + 35 + 24 + 56 + 92 + 16 + 23 x 23 x 45 x 0 ?

Mathematics
1 answer:
8_murik_8 [283]3 years ago
7 0
I will come back to this when I am older
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Write the number that can be shown with this many hundreds ten one hundred 1 ten 2 18 one
noname [10]
1 ten
2 hundreds?
18 ones?

218

5 0
3 years ago
An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than
Sedbober [7]

Testing the hypothesis, it is found that:

a)

The null hypothesis is: H_0: \mu \leq 10

The alternative hypothesis is: H_1: \mu > 10

b)

The critical value is: t_c = 1.74

The decision rule is:

  • If t < 1.74, we <u>do not reject</u> the null hypothesis.
  • If t > 1.74, we <u>reject</u> the null hypothesis.

c)

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of <u>at most 10 pounds</u>, that is:

H_0: \mu \leq 10

At the alternative hypothesis, it is tested if the mean loss is of <u>more than 10 pounds</u>, that is:

H_1: \mu > 10

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a <u>significance level of 0.05</u> and 18 - 1 = <u>17 df.</u>

Hence, using a calculator for the t-distribution, the critical value is: t_c = 1.74.

Hence, the decision rule is:

  • If t < 1.74, we <u>do not reject</u> the null hypothesis.
  • If t > 1.74, we <u>reject</u> the null hypothesis.

Item c:

We have the <u>standard deviation for the sample</u>, hence the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

For this problem, we have that:

\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18

Thus, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}

t = 1.41

Since t = 1.41 < 1.74, we <u>do not reject the null hypothesis</u>, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0
3 years ago
The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly
Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between \\ (\mu - 6\sigma) and \\ (\mu + 6\sigma). It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027

For those with defects <em>operating at a 6 sigma level, </em>the probability is:

\\ 1 - p = 1 - 0.999999998027 = 0.000000001973

Similarly, for finding <em>no defects</em> in a 5 sigma level, we have:

\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697.

The probability of defects is:

\\ 1 - p = 1 - 0.999999426697 = 0.000000573303

Well, the defects present in a six sigma level and a five sigma level are, respectively:

\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}

\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}  

Then, comparing both fractions, we can confirm that a <em>6 sigma level is markedly different when it comes to the number of defects present:</em>

\\ {6\sigma} \approx \frac{2}{10^9} [1]

\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9} [2]

Comparing [1] and [2], a six sigma process has <em>2 defects per billion</em> opportunities, whereas a five sigma process has <em>600 defects per billion</em> opportunities.

8 0
3 years ago
Solve for X. 96 93 6 3
Temka [501]

93 is the right answer please mark me as brainlest

Step-by-step explanation:

I say the answer support. me mark me as brainlest

4 0
2 years ago
Player 1 and player 2 are running laps around a field. Player 1 runs 1 lap every 4 ​minutes, while player 2 runs 1 lap every 15
Nonamiya [84]

Answer:

15 min

Step-by-step explanation:

if it takes Player 1 one min he will meet player two at there second lap after 15 min

4 0
3 years ago
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