#### Overall, there are **two** ways to do this - by adding up several COUNTIF formulas or using a **SUM** COUNTIFS formula with an array constant. Formula 1. Add up **two** or more COUNTIF or COUNITFS formulas. In the table below, supposing you want to count orders with the "Cancelled" and "Pending" status. To have it doen, you can simply write 2 regular. Click on the "Home" tab and expand the "**Number** Format" dropdown. Click on "More **Number** Formats.". Select "Custom," and in the "Type" box, type: [h]: mm;@, and click "OK.". We can see that with this format, the result changes from 03:30 to 27:30. One would save this format in the 'Type' list the next time we need it. Transcript. Example 13 **The sum** of **a two-digit number and** the **number** obtained by reversing the digits is 66. If the digits of the **number** differ by **2**, find the **number**. How many such **numbers** are there? **Number** is of the form Let Digit at Units place = y & Digit at Tens place = x Given that (**Number**) + (Reversed **Number**) = 66 (10x + y) + (10y + x.

**difference**between the

**two numbers**and use that as the basis. So

**difference**of 1 and -1 is

**2**. So 1 contributes (1/

**2**)*100 = 50%. What about -1? Use the absolute function. ( abs(-1) /

**2**) * 100 = 50%. The

**difference**method works fine if you have only

**two numbers**.

**The**

**sum**on

**two**

**numbers**

**is**7 and the

**difference**

**is**1. Find

**number**1 and

**number**2 that make both statements true:

**Number**1 +

**Number**2=7.

**Number**1-Number 2=1. Asked by shavier m #1012493 2 years ago 4/28/2020 10:13 AM. Last updated by natalie p #1073530 a year ago 11/10/2020 2:19 PM. Answers 1.

**The**traditional Fibonacci sequence is 1, 2, 3, 5, 8, 13, 21 and so on, with each

**number**

**the**

**sum**

**of**

**the**preceding

**numbers**. Years ago I began having teams estimate with a modified Fibonacci sequence of 1, 2, 3, 5, 8, 13,

**20**, 40 and 100. Why? It's because

**numbers**that are too close to one another are impossible to distinguish as estimates. Weber.