I will attach google sheet that I used to find regression equation.
We can see that linear fit does work, but the polynomial fit is much better.
We can see that R squared for polynomial fit is higher than R squared for the linear fit. This tells us that polynomials fit approximates our dataset better.
This is the polynomial fit equation:

I used h to denote hours. Our prediction of temperature for the sixth hour would be:

Here is a link to the spreadsheet (
<span>https://docs.google.com/spreadsheets/d/17awPz5U8Kr-ZnAAtastV-bnvoKG5zZyL3rRFC9JqVjM/edit?usp=sharing)</span>
Answer: $106.8
Step-by-step explanation:
Given that the bond price is $100.75
With a 6% brokerage fee
Find the total cost to the nearest cent
First, we calculate the 6% brokerage fee= 6/100 × 100.75
= 0.06 x 100.75
= 6.045
Total cost therefore is the sum of brokerage fee and the cost of bond
= $6.045 + $100.75
= $106.795
To the nearest cent = $106.8
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
Answer:
1:3 is the probability
Step-by-step explanation:
Answer:
16 by 8
Step-by-step explanation: