# Tutor profile: Christi B.

## Questions

### Subject: Microsoft Excel

How can I highlight all the blank cells yellow in my workbook?

The easiest way to do this is to use conditional formatting in the entire workbook. First, select all of the cells in the workbook by clicking the little grey triangle in the upper left corner. Then under the Home tab, click Conditional Formatting and new rule. Then click on "Format only cells that contain" and under the "Format only cells with" option choose Blanks. Then click on the "Format" button, click on the fill tab, select yellow, and then click OK. Click OK once again to finalize your rule and see your results.

### Subject: Pre-Algebra

Jacob is buying candy for his cousin's Christmas present. He buys 10 candy canes for a total of $6.50. How much did each individual candy cane cost?

To solve this problem, we are going to set up a ratio. 10 candy canes = $6.50 1 candy cane =? $$ \frac{$6.50}{10cc} $$ = $$ \frac{$6.50 ÷ 10}{10cc ÷ 10} $$ = $$ \frac{$0.65}{1cc} $$ So 1 candy cane is $0.65.

### Subject: Algebra

In the year 2010, the population of a city was 10 million and increasing continuously at a rate of 3.5% per year. What will the population be in the year 2020?

First, we need to determine what formula we need to use to help us solve this problem. Since we are talking about a starting number, population, and how it is growing over a certain amount of time, we can use the exponential growth rate to help us. $$ A=Pe^{rt}$$ where: A = ending amount P = principal amount, or starting amount r = rate of growth or decay t = time From the problem, we can see that we are given a starting population, a rate, and a time frame. The starting population is 10 million, so we will use P=10, noting that the answer given will be in millions. The rate is 3.5%, so converting to a decimal, we will use r=0.035. The time frame is from 2010 to 2020, which is 10 years, so t=10. Plugging this into our formula, we will get $$A = 10e^{(0.035)(10)}$$. Calculating this, we get an answer of 14.19 million population.

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