Can someone please show the work when doing this !
1 answer:
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Answer:
The variation rate is 4.07 × 10⁻⁵ cm²/ºC
Step-by-step explanation:
The relationship of the change in length as a function of the temperature, which are given in this problem can be written by the expression for the area of a rectangle
a = L × W
Differentiating both sides,
=
= W
+ L
The values they give us are
= 1.1 × 10⁻⁵ cm/ºC
= 8.9 × 10⁻⁶ cm/ºC
W = 1.6 cm
L= 2.6 cm
Substituting the values and calculating
= (1.6 × 1.1 × 10⁻⁵) + (2.6 × 8.9 × 10⁻⁶)
= (1.76 × 10⁻⁵) + (2.31 × 10⁻⁵)
= 4.07 × 10⁻⁵ cm²/ºC
The variation rate is 4.07 × 10⁻⁵ cm²/ºC
Answer:
99% CI: [45.60; 58.00]min
Step-by-step explanation:
Hello!
Your study variable is:
X: Time a customer stays in a certain restaurant. (min)
X~N(μ; σ²)
The population standard distribution is σ= 17 min
Sample n= 50
Sample mean X[bar]= 51.8 min
Sample standard deviation S= 27.68
You are asked to construct a 99% Confidence Interval. Since the variable has a normal distribution and the population variance is known, the statistic to use is the standard normal Z. The formula to construct the interval is:
X[bar] ± *(σ/√n)
Upper level: 51.8 - 2.58*(17/√50) = 45.5972 ≅ 45.60 min
Lower level: 51.8 + 2.58*(17/√50) = 58.0027 ≅58.00 min
With a confidence level of 99%, you'd expect that the interval [45.60; 58.00]min will contain the true value of the average time customers spend in a certain restaurant.
I hope you have a SUPER day!
PS: Missing Data in the attached files.
