Answer:
The critical value for this hypothesis test is 6.571.
Step-by-step explanation:
In this case the professor wants to determine whether the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes.
Then the variance will be, 
The hypothesis to determine whether the population variance is less than 25.0 minutes or not, is:
<em>H</em>₀: The population variance is not less than 25.0 minutes, i.e. <em>σ²</em> = 25.
<em>Hₐ</em>: The population variance is less than 25.0 minutes, i.e. <em>σ²</em> < 25.
The test statistics is:
The decision rule is:
If the calculated value of the test statistic is less than the critical value, 
then the null hypothesis will be rejected.
Compute the critical value as follows:
*Use a chi-square table.
Thus, the critical value for this hypothesis test is 6.571.
x(-1,3) y(3,0) z(-1,-2)
xy
m = (3 + 1)/(0 - 3) = 4/-3 = -4/3
y - 3 = -4/3(x + 1)
y - 3 = -4/3x - 4/3
xy: y = -4/3x + 5/3
yz
m = (0 + 2)/(3 + 1) = 2/4 = 1/2
y = 1/2(x - 3)
yz: y = 1/2x - 3/2
xz
m = (3 + 2)/(-1 + 1) = 5/0 = undefined
y - 3 = undefined(x + 1)
xz: x = -1
xy: y = -4x/3 + 5/3
yz: y = x/2 - 3/2
xz: x = -1
0.636
Add a 0 then a decimal .you will see the number 70 goes with 6 *11=66
then the remainder will be 4 , add a 0 and you will see the number goes with 3*11= 33 then the remainder will be 7 again add a 0 and multiply 6. You will get your answer.
Answer:
12
Step-by-step explanation:
6(d+1)−2d=54
Here, we have to find the value of d
multiply (d+1) by 6
6×d + 6×1 - 2 d= 54
6d + 6 - 2d= 54
Now subtract 2 d from 6d
6d- 2d +6= 54
4 d+ 6=54
Now, by using transposition method,
+ 6 will be transferred from L.H.S to -6 to R.H.S
4 d = 54- 6
4d= 48
Now, divide 48 by 4
d= 48÷ 4
d= 12
Hence, the value of d is 12
Hence, the correct answer is 12
