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amm1812
3 years ago
12

Please answer correctly !!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!

Mathematics
1 answer:
Luba_88 [7]3 years ago
6 0

9514 1404 393

Answer:

  DJ = 48

Step-by-step explanation:

The centroid (J) divides the median into parts that have the ratio 2:1.

  DJ : JH = 2 : 1

  DJ : (DJ +JH) = 2 : (2+1)

  DJ : DH = 2 : 3

  DJ = (2/3)DH = (2/3)(72)

  DJ = 48

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In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean
Nastasia [14]

Answer:

a) \bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

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

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

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

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

b) ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

Step-by-step explanation:

Previous concepts  

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".  

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".  

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n=80 represent the sample size  

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:  

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

For this case we can calculate the mean like this:

\bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

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

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

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

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

Part b

For this case is the sample size is doubled the margin of error would be:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

5 0
2 years ago
Plss help mehhhhhhhhhhh pls
polet [3.4K]

Answer:

Infinitely many solutions.

Step-by-step explanation:

Any value of X makes the statement true, meaning the solution is all real numbers.

8 0
2 years ago
B) A shopkeeper sold her goods for Rs 16,950 allowing 25 % discount and then levied on
Damm [24]

Answer:

Please mark it as the brainliest...

7 0
3 years ago
This is for a grade,Please help me out as much as you can!!!
kolbaska11 [484]

Answer:

a

Step-by-step explanation:

5 0
2 years ago
Bruce is going to call one person from his contacts at random. He has 25 total contacts. 20 of those contacts are from his neigh
PtichkaEL [24]
No of people not from their neighborhood=25-20=5
total contact s=25
therefore probability to call a person not from his neighborhood=5/25=1/5
therefore P=1/5
5 0
3 years ago
Read 2 more answers
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