Answer:
Median is 18, because 18 is the middle number.
Step-by-step explanation:
Answer:
Step-by-step explanation:
9514 1404 393
Answer:
138.77
Step-by-step explanation:
Your scientific or graphing calculator will have exponential functions for bases 10 and e. On the calculator shown in the first attachment, they are shifted (2nd) functions on the log and ln keys. Consult your calculator manual for the use of these functions.
The value can be found using Desmos, the Go.ogle calculator, or any spreadsheet by typing 10^2.1423 as input. (In a spreadsheet, that will need to be =10^2.1423.) The result using the Go.ogle calculator is shown in the second attachment.
You can also use the y^x key or the ^ key (shown to the left of the log key in the first attachment). Again, you would calculate 10^2.1423.
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We have assumed your log is to the base 10. If it is base e (a natural logarithm), then you use the e^x key instead. Desmos, and most spreadsheets, will make use of the EXP( ) function for the purpose of computing e^( ). You can type e^2.1423 into the Go.ogle calculator.
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<em>Additional comment</em>
There are also printed logarithm tables available that you can use to look up the number whose log is 0.1423. You may have to do some interpolation of table values. You should get a value of 1.3877 as the antilog. The characteristic of 2 tells you this value is multiplied by 10^2 = 100 to get the final antilog value.
The logarithm 2.1423 has a "characteristic" (integer part) of 2, and a "mantissa" (fractional part) of 0.1423.
Original price: $3,495
Yearly price decrease = 0.2 x 3,495 = $699 per year
Let y be the value of the car and x be the number of years:
y = 3495 - 699x
After 5 years,
y = 3495 - 699(5)
= $0
In order to solve this problem, we set up a linear equation where the variables were the value of the car and the time passed since purchase. The claim that the car will not be anything is true to an extent because its price has dropped to 0 according to our function; however, practically, a car cannot be $0 so it will be worth at least something, however smaller than its original price.