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Alenkinab [10]
2 years ago
12

When the transmogrifiers are made at the factory, they move along a conveyer belt. Every 8th transmogrifier gets a blue control

panel, every 3rd transmogrifier comes in green, and every 4th transmogrifier comes with a genuine Spaceman Spiff autographed photo. If 150 transmogrifiers come off of the conveyer belt, how many will have a blue control panel, be green, and come with a Spaceman Spiff autographed photo?
Mathematics
2 answers:
Vladimir79 [104]2 years ago
5 0

Answer:

50 green

Step-by-step explanation:

19 blue

38 spaceman

MaRussiya [10]2 years ago
4 0

Answer:

  • Blue - 18, green - 50, with photo - 37

Step-by-step explanation:

<u>Total number = 150</u>

  • Blue = 150/8 = 18 (integer part)
  • Green = 150/3 = 50
  • Spaceman Spiff autograph = 150/4 = 37 (integer part)
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Hatshy [7]

Answer:

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

299.4\leq\mu\leq 330.6

Step-by-step explanation:

We have to construct a 99% confidence interval for the mean.

A sample of n=7 males is taken. We know the sample mean = 315 watts and the sample standard deviation = 16 watts.

For a 99% confidence interval, the value of z is z=2.58.

We can calculate the confidence interval as:

M-z\sigma/\sqrt{n}\leq\mu\leq M+z\sigma/\sqrt{n}\\\\315-2.58*16/\sqrt{7}\leq\mu\leq 315+2.58*16/\sqrt{7}\\\\315-15.6\leq \mu\leq 315+15.6\\\\299.4\leq\mu\leq 330.6

The 99% confidence interval for the mean peak power after training is [299.4, 330.6]

5 0
3 years ago
in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok
grandymaker [24]

Answer:

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

p_v =2*P(Z>1.869)=0.0616  

If we compare the p value obtained and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

\hat p=0.849 estimated proportion of adults were no longer smoking after one month

p_o=0.80 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:  

Null hypothesis:p=0.8  

Alternative hypothesis:p \neq 0.8  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z>1.869)=0.0616  

If we compare the p value obtained and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

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Lady_Fox [76]

Answer:

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Step-by-step explanation:

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Answer:

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