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katen-ka-za [31]
3 years ago
11

AC has endpoints A(3,7) and C(6,11). Find AB if B is the midpoint of AC.

Mathematics
1 answer:
faltersainse [42]3 years ago
8 0

Answer:

d=2.5

Step-by-step explanation:

first find the coordinate of B(mid point of AC):A(3,7) C(6,11)

d=√(6-3)²+(11-7)²

d=√3²+4²

d=√9+16=√25=5

since B is the mid point : d/2=5/2=2.5

<h2>Another way :</h2>

B(x1+x2/2 , y1+y2/2)   , x1=3 , x2=6, y1=7, y2=11

B(9/2,18/2)

B(9/2,9)

Find AB : the length or distance between 2 points:

d=√(x2-x1)²+(y2-y1)²

d=√(3-9/2)²+(7-9)²

d=√(-3/2)²+(-2)²

d=√1.5²+4

d=√6.25

d=2.5

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what is the relationship between 5.34 x 10^5 and 5.34 x 10^-2 is it greater or lesser than? for 5.34 x 10^5? will give brainlies
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the relationship between 5.34 x 10^5 and  5.34 x 10^-2

5.34 x 10^5 = 5.34*100000= 534000

5.34 * 10^-2  , for exponent -2 we move the decimal point 2 places to the left

5.34 x 10^-2= 0.0534

Now we compare 534000  and 0.0534

0.0534 is 10,000,000 times less than 534000

Or we can say  534000 is 10,000,000 times greater than 0.0534

5.34 x 10^5 is 10,000,000 times greater than 5.34 x 10^-2


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The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
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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.

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