5 pounds for 6.65 because you will get more pounds for a good amount of money.
Let X be the national sat score. X follows normal distribution with mean μ =1028, standard deviation σ = 92
The 90th percentile score is nothing but the x value for which area below x is 90%.
To find 90th percentile we will find find z score such that probability below z is 0.9
P(Z <z) = 0.9
Using excel function to find z score corresponding to probability 0.9 is
z = NORM.S.INV(0.9) = 1.28
z =1.28
Now convert z score into x value using the formula
x = z *σ + μ
x = 1.28 * 92 + 1028
x = 1145.76
The 90th percentile score value is 1145.76
The probability that randomly selected score exceeds 1200 is
P(X > 1200)
Z score corresponding to x=1200 is
z =
z =
z = 1.8695 ~ 1.87
P(Z > 1.87 ) = 1 - P(Z < 1.87)
Using z-score table to find probability z < 1.87
P(Z < 1.87) = 0.9693
P(Z > 1.87) = 1 - 0.9693
P(Z > 1.87) = 0.0307
The probability that a randomly selected score exceeds 1200 is 0.0307
Answer:
The Proof and Explanation for
Part C ,
Qs 9 and
Qs 10 are below.
Step-by-step explanation:
PART C .
Given:
AD || BC ,
To Prove:
ΔAED ≅ ΔCEB
Proof:
Statement Reason
1. AD || BC 1. Given
2. ∠A ≅ ∠C 2. Alternate Angles Theorem as AD || BC
3. ∠AED ≅ ∠CEB 3. Vertical Opposite Angle Theorem.
4. AE ≅ EC 4. Given
5. ΔAED ≅ ΔCEB 5. By A-S-A congruence test....Proved
Qs 9)
Given:
AB ≅ BC ,
∠ABD ≅ ∠CBD
To Prove:
∠A ≅ ∠C
Proof:
Statement Reason
1. ∠ABD ≅ ∠CBD 1. Given
2. AB ≅ CB 2. Given
3. BD ≅ BD 3. Reflexive Property
4. ΔABD ≅ ΔCBD 4. By S-A-S congruence test
5. ∠A ≅ ∠C 5. Corresponding parts of congruent Triangles Proved.
Qs 10)
Given:
∠MCI ≅ ∠AIC
MC ≅ AI
To Prove:
ΔMCI ≅ ΔAIC
Proof:
Statement Reason
1. ∠MCI ≅ ∠AIC 1. Given
2. MC ≅ AI 2. Given
3. CI ≅ CI 3. Reflexive Property
4. ΔMCI ≅ ΔAIC 4. By S-A-S congruence test