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trasher [3.6K]
3 years ago
7

The shape on the left is translated to the shape on the right which statement about transformation is true?

Mathematics
1 answer:
Oksanka [162]3 years ago
3 0
When. Dbhdbhsbbsb send Sheba ebbs she be be she ehehhd rhehebjebd jehd Ivan’s shhdbdhs ehuehe while she dhhe e hehebehe
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9. the learning modalityyou prefer this new normal a.modular learning b.online learning c.digital learning​
Gwar [14]

Answer: A module learning

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
MAXImum [283]

Answer:

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d.  The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

t(0.01)(49)= 1.299

t(0.05)= 1.677  

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0  for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test  we reject H0   as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ±  z∝/2  (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0
3 years ago
What is... 15% of x Please help
lukranit [14]
This answer will probably be 3/20 x not sure
7 0
3 years ago
!!!PLEASE HELPP ME!!
Annette [7]

Answer:

negative :)

Step-by-step explanation:

hope this helps

6 0
2 years ago
Read 2 more answers
6c to te second power - 5d + 8
mina [271]
Lets see,
6c to the second power is 36c because 6c times 6c is 36 c
And since the variables are different and only one number doesn't have a variable, 
The answer would be, 36c-5d+8
7 0
3 years ago
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