<span>4.5 + 0 = 3.5 + 0.7x
1 = 0.7x
x = 1/0.7
x = 1.43 gallons of distilled water
Hope this helps!!</span>
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
Answer:
3- 0.14 2- 841
Step-by-step explanation:
3- 0.7*0.2=0.14
2- 56/2= 28
28*28+29*29= 841*841
square root of 841*841= 841
Answer:
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
Step-by-step explanation:
Here in this question, we want to state what will happen if the null hypothesis is true in a chi-square test.
If the null hypothesis is true in a chi-square test, discrepancies between observed and expected frequencies will tend to be small enough to qualify as a common outcome.
This is because at a higher level of discrepancies, there will be a strong evidence against the null. This means that it will be rare to find discrepancies if null was true.
In the question however, since the null is true, the discrepancies we will be expecting will thus be small and common.
Answer:
1. 15 - 5n where n>=1
2. n² where n>=1
Step-by-step explanation:
1. {10, 5, 0, -5, -10} is an Arithmetic Progression
nth term is a + (n - 1)d
where a = first term, n= nth term, d= common difference.
a = 10, d = -5 (5-10, 0-5, -5-0, -10-(-5))
Therefore, nth(General) term of the sequence:
= 10 + (n - 1)-5
= 10 + (-5n) + 5
= 10 + 5 - 5n
= 15 - 5n
Test:
if n = 1; 15 - 5(1) = 10
if n = 2; 15 - 5(2) = 5
if n = 3; 15 - 5(3) = 0 and so on.
2. {1, 4, 9, 16, 25}
The general term of the sequence is n²
Test:
if n = 1; 1² = 1
if n = 2; 2² = 4
if n = 3; 3² = 9 and so on.
