Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation
Calculate z-scores for the following random variable and determine their probabilities from standard tables.
x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088
x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912
x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587
x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413
Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36
Answer: 27
Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36
Answer: 27
Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78
Answer: 204
Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150
Answer: 150
Answer:- B. No, because the corresponding congruent angles listed are not the included angles.
Explanation:-
Given:- ΔWXY and ΔBCD with ∠X ≅∠C, WX ≅ BC, and WY ≅ BD.
Now, look at the attachment
We can see that ∠X and ∠C are not included angles by the corresponding equal sides.
⇒ We cannot use SAS postulate to show ΔWXY ≅ ΔBCD .
⇒ B is the right option.
SAS postulate tells the if two sides of a triangle and their included angle is equal to the two sides of a triangle and their included angle of another triangle then the two triangles are congruent.
So increased weight = 92-80 = 12
so ℅ = 12×100/80 = 15
so his weight is increased by 15℅ !!
The answer is <span>-1.2917960675</span>
Step-by-step explanation:
B. Is correct, the graph will never touch 0 if we shift the graphs horizontally,