Formula for this case is
(a+b)∧3=a∧3+3a∧2b+3ab∧2+b∧3
(2x+5)∧3=(2x)∧3+3*(2x)∧2*5+3*2x*5∧2+5∧3=8 x∧3+3*4x∧2*5+3*2x*25+125=
=8x∧3+60x∧2+150x+125
Good luck!!!
Answer:
I think that the correct answer is B. 29k + 4
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
2.5 weighs more. It’s a greater number than 2.48
Answer:
Dimensions of the rectangular plot will be 500 ft by 750 ft.
Step-by-step explanation:
Let the length of the rectangular plot = x ft.
and the width of the plot = y ft.
Cost to fence the length at the cost $3.00 per feet = 3x
Cost to fence the width of the cost $2.00 per feet = 2y
Total cost to fence all sides of rectangular plot = 2(3x + 2y)
2(3x + 2y) = 6,000
3x + 2y = 3,000 ----------(1)
3x + 2y = 3000
2y = 3000 - 3x
y = ![\frac{1}{2}[3000-3x]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B3000-3x%5D)
y = 1500 - 
Now area of the rectangle A = xy square feet
A = x[
]
For maximum area 
A' =
= 0
1500 - 3x = 0
3x = 1500
x = 500 ft
From equation (1),
y = 1500 - 
y = 1500 - 750
y = 750 ft
Therefore, for the maximum area of the rectangular plot will be 500 ft × 750 ft.
two fencing 3(500+500) = $3000
other two fencing 2(750+750) = $3000