K=4(5)
k=20
you just plug in 5 for d and multiply normally from there.
Answer:
Since the calculated value of t =-0.427 does not fall in the critical region so we accept H0 and conclude that there is enough evidence to show that mean difference in the age of onset of symptoms and age of diagnosis is 25 months .
Step-by-step explanation:
The given data is
Difference d= -24 -12 -55 -15 -30 -60 -14 -21 -48 -12 -25 -53 -61 -69 -80
∑ d= -579
∑d²= 29871
1) Let the hypotheses be
H0: ud= 25 against the claim Ha: ud ≠25
H0 : mean difference in the age of onset of symptoms and age of diagnosis is 25 months .
Ha: mean difference in the age of onset of symptoms and age of diagnosis is not 25 months.
2) The degrees of freedom = n-1= 15-1= 14
3) The significance level is 0.05
4) The test statistic is
t= d`/sd/√n
The critical region is ║t║≤ t (0.025,14) = ±2.145
d`= ∑di/n= -579/15= -38.6
Sd= 23.178 (using calculators)
Therefore
t= d`/ sd/√n
t= -38.6/ 23.178√15
t= -1.655/3.872= -0.427
5) Since the calculated value of t =-0.427 does not fall in the critical region so we accept H0 and conclude that there is enough evidence to show that mean difference in the age of onset of symptoms and age of diagnosis is 25 months .
I actually believe that the answer is A
Answers:
Part A: 12y² + 10y – 21
Part B: 4y³ + 6y² + 6y – 5
Part C: See below.
Explanations:
Part A:
For this part, you add Sides 1, 2 and 3 together by combining like terms:
Side 1 = 3y² + 2y – 6
Side 2 = 4y² + 3y – 7
Side 3 = 5y² + 5y – 8
3y² + 2y – 6 + 4y² + 3y – 7 + 5y² + 5y – 8
Combine like terms:
3y² + 4y² + 5y² + 2y + 3y + 5y – 6 – 7 – 8
12y² + 10y – 21
Part B:
You have the total perimeter and the sum of three of the sides, so you just need that fourth side value, which we can call d.
P = 4y³ + 18y² + 16y – 26
Sides 1, 2 & 3 = 12y² + 10y – 21
Create an algebraic expression:
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
Solve for d:
12y² + 10y – 21 + d = 4y³ + 18y² + 16y – 26
– 12y² – 12y²
10y – 21 + d = 4y³ + 6y² + 16y – 26
– 10y – 10y
– 21 + d = 4y³ + 6y² + 6y – 26
+ 21 + 21
d = 4y³ + 6y² + 6y – 5
Part C:
If closed means that the degree that these polynomials are at stay that way, then yes, this is true in these cases because you will notice that each side had a y², y and no coefficient value except for the fourth one. This didn't change, because you only add and subtract like terms.
