Answer:
t=5.5080( to 3 d.p)
Step-by-step explanation:
From the data given,
n =20
Deviation= 34/20= 1.7
Standard deviation (sd)= 1.3803(√Deviation)
Standard Error = sd/√n
= 1.3803/V20 = 0.3086
Test statistic is:
t = deviation /SE
= 1.7/0.3086 = 5.5080
ndf = 20 - 1 = 19
alpha = 0.01
One Tailed - Right Side Test
From Table, critical value of t =2.5395
Since the calculated value of t = 5.5080 is greater than critical value of t = 2.5395, the difference is significant. Reject null hypothesis.
t score = 5.5080
ndf = 19
One Tail - Right side Test
By Technology, p - value = 0.000
Since p - value is less than alpha , reject null hypothesis.
Conclusion:
From the result obtained it can be concluded that ,the data support the claim that the mean rating assigned to the wine when the cost is described as $90 is greater than the mean rating assigned to the wine when the cost is described as $10.
So the distance formula: d = rad (x2 - x1)^2 + (y2 - y1)^2
it doesn't matter what order you do, you just have to make sure it's a y coordinate for y and an x coordinate for x.
rad (9-5)^2 + ((-6)-1)^2
rad 4^2 + (-7)^2
rad 16 + 49
rad 65
Answer:
b
Step-by-step explanation:
A+b+c = 131
b = 7 + 2a and c = a - 12
so a + 7+2a + a-12 = 131
then 4a -5 = 131
then 4a = 136
and a = 34....so b= 75 and c = 22....check 34+75+22=131√
Answer:
Sum of 
and 
is 
Option B is correct answer.
Step-by-step explanation:
We need to find sum of and
Finding sum of 
and :
We know that 
Replacing x^2-4
Now, taking LCM of (x+2)(x-2) and (x+2) we get (x+2)(x-2)
So, Sum of and is
Option B is correct answer.
