The equation to show the depreciation at the end of x years is
Data;
- cost of machine = 1500
- annual depreciation value = x
<h3>Linear Equation</h3>
This is an equation written to represent a word problem into mathematical statement and this is easier to solve.
To write a linear depreciation model for this machine would be
For number of years, the cost of the machine would become
This is properly written as
where x represents the number of years.
For example, after 5 years, the value of the machine would become
The value of the machine would be $500 at the end of the fifth year.
From the above, the equation to show the depreciation at the end of x years is f(x) = 1500 - 200x
Learn more on linear equations here;
brainly.com/question/4074386
Answer:
y = lnx / ln 7.
Step-by-step explanation:
x = 7^y
Take logarithms of both sides:
ln x = ln 7^y
ln x = y ln 7 ( because ln a^b = b ln a. One of the Laws of Logarithms).
y = lnx / ln 7.
Answer:
The store can be 90% confident that the interval from 0.293 to 0.355 captures the proportion of all customers of this store who have moved in the past 5 years.
Step-by-step explanation:
The confidence interval provides a range of values (lower and upper bound) based on a certain confidence level at which the true proportion or mean value of a given sample mean or proportion exists. In the scenario above, the confidence level is 90% and the confidence interval is 0.293 to 0.355. Hence, we can be 90% confident that the true proportion of all customers of the store who have moved within the last five years exists within this interval.
Answer: Our required interval for temperature is 
Step-by-step explanation:
Since we have given that
The normal temperature range for Yuma, on Janurary is atleast = 48 degrees
and the highest temperature on that day is no higher than 64 degrees.
Let the temperature in Yuma be 'x'.
So, we know that for atleast means it can 48 degrees or more, and no higher than means it should be equa to or less than 64 degrees.
Mathematically, it is expressed as
Hence, our required interval for temperature is
