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skad [1K]
4 years ago
7

Please help me with this question and explain thankss

Mathematics
1 answer:
butalik [34]4 years ago
6 0
∠BDC = ∠BEC = 40°
∠ADB = (arc AB)/2 = 44°/2 = 22°
∠ADC = ∠ADB +∠BDC
  = 22° +40° = 62°
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Commute times in the U.S. are heavily skewed to the right. We select a random sample of 45 people from the 2000 U.S. Census who
Delicious77 [7]

Answer:

The mean commute time in the U.S. is less than half an hour.

Step-by-step explanation:

In this case we need to test whether the mean commute time in the U.S. is less than half an hour.

The information provided is:

 n=45\\\bar x=25.5\\s=19.1\\\alpha =0.05

(a)

The hypothesis for the test can be defined as follows:

<em>H</em>₀: The mean commute time in the U.S. is not less than half an hour, i.e. <em>μ</em> ≥ 30.

<em>Hₐ</em>: The mean commute time in the U.S. is less than half an hour, i.e. <em>μ</em> < 30.

(b)

As the population standard deviation is not known we will use a t-test for single mean.

Compute the test statistic value as follows:

 t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{25.2-30}{19.1/\sqrt{45}}=-1.58

Thus, the test statistic value is -1.58.

(c)

Compute the p-value of the test as follows:

p-value=P(t_{(n-1)}  

*Use a t-table.

The p-value of the test is 0.061.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

p-value = 0.061> α = 0.05

The null hypothesis will not be rejected at 5% level of significance.

Thus, concluding that the mean commute time in the U.S. is less than half an hour.

8 0
4 years ago
Wats slope please hurry!!!!!! please!!??<br><br> 2<br><br> 1<br><br> 5
Dennis_Churaev [7]
The slope is 1 :)

ajndjshdbsusjdbsjidbs
7 0
3 years ago
How would I convert 2.7 g/cm to lb/in?
ozzi
It will be 0.0151193
hope that helped
4 0
3 years ago
Suppose you know the length of a confidence interval of a population mean is 8.4 and the sample mean (x bar) is 10. Find the ​ma
SOVA2 [1]

Answer:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X=10 represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Solution to the problem

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The sample mean \bar X is distributed on this way:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})  

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

The margin of error is given by:

ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The lenght of the interval correspond to:

8.4 = 2ME

ME= \frac{8.4}{2}= 4.2

And since we know the margin of error we can find the limits for the confidence interval:

Lower = 10 -4.2=5.8

Upper = 10 +4.2=14.2

4 0
3 years ago
Please help asap and explain!! i give brainliests!!
Snezhnost [94]
Change of y over change of x
2.5 - 1.25 = 1.25
4 - 2 = 2
1.25/2 = .625
Hope this helps!
The answer is .625 if theres any confusion 
8 0
3 years ago
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