Answer:
Length = 49.5 unit and width = 49.5 unit
Step-by-step explanation:
Given as , Perimeter of rectangle = 198 unit
so ,as Perimeter of rectangle = 2× ( Length + width)
Or, 198 = 2 × (Length + width)
Or, 
= length + width
So, length + width = 99 unit
Now to make area maximum
Length × width = maximum
Or, (99 - width ) × width = maximum
99 Width - width² = maximum Let width = W
Now differentiate both side with respect to W
D(99W - W²)
= 0 as, constant diff is 0
So, 99 - 2w = 0
Or, w = 
Or, w = 49.5 unit and L = 99- 4905 = 49.5 unit Answer
Answer:
Think its 3 but don't enter it, wait for someone else
Answer: When humans urbanize an area they
Step-by-step explanation: I think this because before people urbanized the place it was probably like a rural place or suburban. When you urbanize a place you might be taking a person or an animals homeland away and that can be a bad impact on them.
Answer:
3
Step-by-step explanation:
Evaluate x/4 + 6 (x - 12) where x = 12:
x/4 + 6 (x - 12) = 12/4 + 6 (12 - 12)
Hint: | Reduce 12/4 to lowest terms. Start by finding the GCD of 12 and 4.
The gcd of 12 and 4 is 4, so 12/4 = (4×3)/(4×1) = 4/4×3 = 3:
3 + 6 (12 - 12)
Hint: | Look for the difference of two identical terms.
12 - 12 = 0:
6×0 + 3
Hint: | Any number times zero is zero.
0×6 = 0:
0 + 3
Hint: | Simplify the expression.
Write 3 + 0 as 3:
Answer: 3
Complete question:
He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is
a) less than 8 minutes
b) between 8 and 9 minutes
c) less than 7.5 minutes
Answer:
a) 0.0708
b) 0.9291
c) 0.0000
Step-by-step explanation:
Given:
n = 47
u = 8.3 mins
s.d = 1.4 mins
a) Less than 8 minutes:
P(X' < 8) = P(Z< - 1.47)
Using the normal distribution table:
NORMSDIST(-1.47)
= 0.0708
b) between 8 and 9 minutes:
P(8< X' <9) =![[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B8-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%3C%20%5Cfrac%7BX%27-u%7D%7Bs.d%2F%20%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B9-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D)
= P(-1.47 <Z< 6.366)
= P( Z< 6.366) - P(Z< -1.47)
Using normal distribution table,
0.9999 - 0.0708
= 0.9291
c) Less than 7.5 minutes:
P(X'<7.5) = ![P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]](https://tex.z-dn.net/?f=%20P%20%5BZ%3C%20%5Cfrac%7B7.5-8.3%7D%7B1.4%2F%20%5Csqrt%7B47%7D%7D%5D%20)
P(X' < 7.5) = P(Z< -3.92)
NORMSDIST (-3.92)
= 0.0000
