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cricket20 [7]
3 years ago
7

The lieutenant noted that 0.68 of the frontiersmen at the conclave wore buckskin. If 512 did not wear buckskin, how many frontie

rsmen were at the conclave?
I need the answer now ASAP!!!!!!!!
Mathematics
2 answers:
Lubov Fominskaja [6]3 years ago
6 0
1600 is the answer. BRANLIEST? :)
pogonyaev3 years ago
4 0

Answer:

1600.

Step-by-step explanation:

The fraction who did not wear buckskin

= 1-0.68 = 0.32.

So 0.32x = 512

x = 51/.0.32 = 1600.

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2x – 1+ 3x=1+ 2x - 5<br> I need the steps
Anettt [7]

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x = -5/3

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3 0
3 years ago
The mean weight of 10 randomly selected newborn babies at a local hospital is 7.14 lbs and the standard deviation is 0.87 lbs. A
sergejj [24]

Answer:

a) ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035  

b) 90% of all samples of size 10 have sample means within 0.5035 of the population mean.

c) The 90% confidence interval would be given by (6.636;7.643)    

d) Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

e) If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value t_{\alpha/2} increase. So then the new interval would be wider than the original.

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=7.14 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=0.87 represent the sample standard deviation

n=10 represent the sample size  

Part a

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And the margin of error is given by:

ME= t_{\alpha/2}\frac{s}{\sqrt{n}}  

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=10-1=9

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that t_{\alpha/2}=1.83

ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035  

Part b

90% of all samples of size 10 have sample means within 0.5035 of the population mean.

Part c

Now we have everything in order to replace into formula (1):

7.14-1.83\frac{0.87}{\sqrt{10}}=6.637    

7.14+1.83\frac{0.87}{\sqrt{10}}=7.643    

So on this case the 90% confidence interval would be given by (6.636;7.643)    

Part d

Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

Part e

If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value t_{\alpha/2} increase. So then the new interval would be wider than the original.

7 0
4 years ago
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