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Andreas93 [3]
3 years ago
13

The regular hexagon ABCDEF rotates 240º counterclockwise about its center to form hexagon A′B′C′D′E′F′. Point C′ of the image co

incides with point A, B, E OR F of the preimage. Point D′ of the image coincides with point A, B, E OR F of the preimage.
Mathematics
2 answers:
Mila [183]3 years ago
4 0

The regular hexagon ABCDEF rotates 240º counterclockwise about its center to form hexagon A′B′C′D′E′F′. Point C′ of the image coincides with point _E_ of the preimage. Point D′ of the image coincides with point _F_ of the preimage.

juin [17]3 years ago
4 0
Jensei ndaklncdnklmpdvsjekekeken
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alex rents a car for one day. the charge is $18 plus $0.12 per mile. alex wants to spend exactly $30. how many miles can he driv
BartSMP [9]

$100 miles can alex drive.

<h3>How to find the how many miles can alex drive?</h3>

Alex rents a car for one day.

The charge is $18 plus $0.12 per mile.

Alex wants to spend exactly $30.

now find how many miles can alex drive as follows below process.

$0.12 chagre for 1 mile and $18 for car rent.

so, subtract the car rent in $30 = $30-$18 = $12

let Alex would drive x miles.

0.12(x) = 12

x = 12/0.12

x = 100

so, $100 miles can alex drive.

Learn more about algebraic, refer:

https://brainly.in/question/610053

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2 years ago
This table of values represents a linear function.
Vesna [10]
4.5 to 6.5 i think not really shure

3 0
3 years ago
Please help me in a rush
vova2212 [387]

5.89 x 5 = 29.45

Hope this helps you!

Always remember you are a Work Of Art!

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3 0
4 years ago
Read 2 more answers
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
4 − 3( − 2) + (1/2)(6 − 8)
geniusboy [140]

Answer:

9

Step-by-step explanation:

8 0
3 years ago
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