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kvv77 [185]
3 years ago
6

Check this please I need grade

Mathematics
1 answer:
Elden [556K]3 years ago
3 0
All good and for number 6 you should draw an acute angle
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If I have a 62% in a class, what percentage do I need to get on a 50% weighted test grade to have at least a 70% in the class?
Nastasia [14]

Answer:

78%

Step-by-step explanation:

70 = 0.50 (62) + 0.50 (x)

70 = 31 + 0.5x

39 = 0.5x

x = 78

You must get at least a 78% on the test to have at least 70% in the class.

3 0
3 years ago
Which of the following is not a property of a rectangle
In-s [12.5K]

Answer:

I would like to say A because Diagonals of a rectangle Bisect each other at a right angle.

This seems likely

Please Mark as Brainliest

Hope this Helps

5 0
3 years ago
Please help NO LINKS ( i will mark brainliest)
Arlecino [84]

Answer:

with what

Step-by-step explanation:

5 0
3 years ago
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
On a particular day, the wind added 4 miles per hour to Alfonso’s rate when he was cycling with the wind and subtracted 4 miles
Greeley [361]

d = r · t ⇒ t = d/r

going

t = 54/(r + 4)

coming

t = 30/(r - 4)

54/(r + 4) = 30/(r - 4) times are equal

54(r - 4) = 30(r + 4) product means/extremes

54r - 216 = 30r + 120 distribute

24r - 216 = 120 subtract 30r from both sides

24r = 336 add 216 to both sides

r = 14 divide both sides by 24

Alfonso bikes at 24 mph.

8 0
3 years ago
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