diff --git a/02_activities/assignments/Assignment2.md b/02_activities/assignments/Assignment2.md
index 05244623b..2abe69060 100644
--- a/02_activities/assignments/Assignment2.md
+++ b/02_activities/assignments/Assignment2.md
@@ -53,8 +53,9 @@ The store wants to keep customer addresses. Propose two architectures for the CU
 
 **HINT:** search type 1 vs type 2 slowly changing dimensions. 
 
-```
-Your answer...
+By adding a "customer address" table with three columns, we can:
+1- use an FK to connect with customer_id from the "customer" table, thus overwriting if needed.
+2- store old and new customer addresses in new columns.
 ```
 
 ***
diff --git a/02_activities/assignments/assigment1_data_model.jpg b/02_activities/assignments/assigment1_data_model.jpg
new file mode 100644
index 000000000..440009ee2
Binary files /dev/null and b/02_activities/assignments/assigment1_data_model.jpg differ
diff --git a/02_activities/assignments/assignment1.sql b/02_activities/assignments/assignment1.sql
index 2e89fa7af..92790fe7a 100644
--- a/02_activities/assignments/assignment1.sql
+++ b/02_activities/assignments/assignment1.sql
@@ -4,21 +4,29 @@
 
 --SELECT
 /* 1. Write a query that returns everything in the customer table. */
-
-
+SELECT * FROM customer;
 
 /* 2. Write a query that displays all of the columns and 10 rows from the cus- tomer table, 
 sorted by customer_last_name, then customer_first_ name. */
 
-
+SELECT * FROM customer
+ORDER BY customer_last_name, customer_first_name
+LIMIT 10;
 
 --WHERE
 /* 1. Write a query that returns all customer purchases of product IDs 4 and 9. */
 -- option 1
-
+SELECT  *
+FROM customer_purchases
+WHERE 
+    product_id = 4 or product_id = 9;
+	
 
 -- option 2
-
+SELECT  *
+FROM customer_purchases
+WHERE 
+    product_id IN (4, 9);
 
 
 /*2. Write a query that returns all customer purchases and a new calculated column 'price' (quantity * cost_to_customer_per_qty), 
@@ -27,10 +35,22 @@ filtered by vendor IDs between 8 and 10 (inclusive) using either:
 	2.  one condition using BETWEEN
 */
 -- option 1
+SELECT  *
+,(quantity * cost_to_customer_per_qty) AS price
+FROM 
+    customer_purchases cp
+WHERE 
+    cp.vendor_id >= 8 
+    AND cp.vendor_id <= 10;
 
 
 -- option 2
-
+SELECT  *
+,(quantity * cost_to_customer_per_qty) AS price
+FROM 
+    customer_purchases cp
+WHERE 
+    cp.vendor_id BETWEEN 8 AND 10;
 
 
 --CASE
@@ -39,19 +59,45 @@ Using the product table, write a query that outputs the product_id and product_n
 columns and add a column called prod_qty_type_condensed that displays the word “unit” 
 if the product_qty_type is “unit,” and otherwise displays the word “bulk.” */
 
+SELECT product_id, product_name
+,CASE 
+When product_qty_type = 'unit' then 'unit'
+When product_qty_type in ('lbs', 'oz')  then 'bulk'
+When product_qty_type is null THEN NULL
+ELSE NULL
+END as prod_qty_type_condensed
+
+FROM product
 
 
 /* 2. We want to flag all of the different types of pepper products that are sold at the market. 
 add a column to the previous query called pepper_flag that outputs a 1 if the product_name 
 contains the word “pepper” (regardless of capitalization), and otherwise outputs 0. */
 
-
+SELECT product_id, product_name
+,CASE 
+When product_qty_type = 'unit' then 'unit'
+When product_qty_type in ('lbs', 'oz')  then 'bulk'
+When product_qty_type is null THEN NULL
+ELSE NULL
+END as prod_qty_type_condensed
+, CASE
+when lower (product_name) like '%pepper%' THEN 1
+ELSE 0
+END AS PEPPER_FLAD
+FROM product
 
 --JOIN
 /* 1. Write a query that INNER JOINs the vendor table to the vendor_booth_assignments table on the 
 vendor_id field they both have in common, and sorts the result by vendor_name, then market_date. */
 
-
+SELECT *
+FROM 
+    vendor v
+INNER JOIN 
+    vendor_booth_assignments vba ON v.vendor_id = vba.vendor_id
+ORDER BY 
+    v.vendor_name, vba.market_date;
 
 
 /* SECTION 3 */
@@ -60,7 +106,12 @@ vendor_id field they both have in common, and sorts the result by vendor_name, t
 /* 1. Write a query that determines how many times each vendor has rented a booth 
 at the farmer’s market by counting the vendor booth assignments per vendor_id. */
 
-
+SELECT 
+    vendor_id
+	,COUNT(*) AS booth_rented
+FROM 
+vendor_booth_assignments
+GROUP by 1;
 
 /* 2. The Farmer’s Market Customer Appreciation Committee wants to give a bumper 
 sticker to everyone who has ever spent more than $2000 at the market. Write a query that generates a list 
@@ -68,6 +119,22 @@ of customers for them to give stickers to, sorted by last name, then first name.
 
 HINT: This query requires you to join two tables, use an aggregate function, and use the HAVING keyword. */
 
+SELECT 
+c.customer_first_name
+,c.customer_last_name
+
+FROM (
+SELECT customer_id
+, sum ((quantity * cost_to_customer_per_qty)) as total_spent
+
+FROM customer_purchases
+GROUP by 1
+HAVING total_spent > 2000
+) TS
+
+left JOIN customer c
+on ts.customer_id = c.customer_id
+ORDER by customer_first_name, customer_last_name;
 
 
 --Temp Table
@@ -82,7 +149,16 @@ When inserting the new vendor, you need to appropriately align the columns to be
 VALUES(col1,col2,col3,col4,col5) 
 */
 
+CREATE TEMP TABLE new_vendor AS
+
+SELECT *
+from vendor;
+
+INSERT into new_vendor
+VALUES (10, "Thomass Superfood store", "'a Fresh Focused store", "Thomas",  "Rosenthal");
 
+SELECT *
+FROM new_vendor;
 
 -- Date
 /*1. Get the customer_id, month, and year (in separate columns) of every purchase in the customer_purchases table.
@@ -90,6 +166,11 @@ VALUES(col1,col2,col3,col4,col5)
 HINT: you might need to search for strfrtime modifers sqlite on the web to know what the modifers for month 
 and year are! */
 
+SELECT
+customer_id
+, strftime ('%m', market_date) as month
+, strftime ('%Y', market_date) as year
+from customer_purchases;
 
 
 /* 2. Using the previous query as a base, determine how much money each customer spent in April 2022. 
@@ -98,3 +179,18 @@ Remember that money spent is quantity*cost_to_customer_per_qty.
 HINTS: you will need to AGGREGATE, GROUP BY, and filter...
 but remember, STRFTIME returns a STRING for your WHERE statement!! */
 
+SELECT
+customer_id
+, month
+, year
+, sum (price) as monthly_spent
+
+FROM (
+SELECT customer_id
+, strftime ('%m', market_date) as month
+, strftime ('%Y', market_date) as year
+, (quantity * cost_to_customer_per_qty) as price
+from customer_purchases
+) tl
+
+GROUP  by 1,2,3;
\ No newline at end of file
diff --git a/02_activities/assignments/assignment2.sql b/02_activities/assignments/assignment2.sql
index 5ad40748a..9bef46194 100644
--- a/02_activities/assignments/assignment2.sql
+++ b/02_activities/assignments/assignment2.sql
@@ -21,6 +21,16 @@ Edit the appropriate columns -- you're making two edits -- and the NULL rows wil
 All the other rows will remain the same.) */
 
 
+SELECT 
+product_name || ', ' || product_size|| ' (' || product_qty_type || ')'
+FROM product;
+SELECT
+    product_name || ', ' ||
+    COALESCE(product_size, '') || ' (' ||
+    COALESCE(product_qty_type, 'unit') || ')' AS detailed_product_info
+FROM product;
+
+
 
 --Windowed Functions
 /* 1. Write a query that selects from the customer_purchases table and numbers each customer’s  
@@ -32,6 +42,16 @@ each new market date for each customer, or select only the unique market dates p
 (without purchase details) and number those visits. 
 HINT: One of these approaches uses ROW_NUMBER() and one uses DENSE_RANK(). */
 
+SELECT
+    customer_id,
+    market_date,
+    DENSE_RANK() OVER (PARTITION BY customer_id ORDER BY market_date) AS visit_number
+FROM
+    customer_purchases
+GROUP BY
+    customer_id,
+    market_date;
+
 
 
 /* 2. Reverse the numbering of the query from a part so each customer’s most recent visit is labeled 1, 
@@ -39,11 +59,29 @@ then write another query that uses this one as a subquery (or temp table) and fi
 only the customer’s most recent visit. */
 
 
+WITH CustomerVisits AS (
+  SELECT
+      customer_id,
+      market_date,
+      ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY market_date DESC) AS visit_number
+  FROM
+      customer_purchases
+)
+SELECT
+    customer_id,
+    market_date
+FROM
+    CustomerVisits
+WHERE
+    visit_number = 1;
 
 /* 3. Using a COUNT() window function, include a value along with each row of the 
 customer_purchases table that indicates how many different times that customer has purchased that product_id. */
 
-
+SELECT *
+,COUNT(*) OVER (PARTITION BY customer_id, product_id) AS product_purchase_count
+FROM
+    customer_purchases;
 
 -- String manipulations
 /* 1. Some product names in the product table have descriptions like "Jar" or "Organic". 
@@ -57,10 +95,18 @@ Remove any trailing or leading whitespaces. Don't just use a case statement for
 
 Hint: you might need to use INSTR(product_name,'-') to find the hyphens. INSTR will help split the column. */
 
+SELECT
+    product_name,
+    TRIM(SUBSTR(product_name, INSTR(product_name, '-') + 1)) AS description
+FROM
+    product;
 
 
 /* 2. Filter the query to show any product_size value that contain a number with REGEXP. */
 
+SELECT product_name, product_size
+FROM product
+WHERE product_size REGEXP '[0-9]';
 
 
 -- UNION
@@ -74,6 +120,34 @@ HINT: There are a possibly a few ways to do this query, but if you're struggling
 with a UNION binding them. */
 
 
+WITH SalesByDate AS (
+    SELECT
+        market_date,
+        SUM(quantity * cost_to_customer_per_qty) AS total_sales
+    FROM customer_purchases
+    GROUP BY market_date
+),
+RankedSales AS (
+    SELECT
+        market_date,
+        total_sales,
+        RANK() OVER (ORDER BY total_sales DESC) AS rank_highest,
+        RANK() OVER (ORDER BY total_sales ASC) AS rank_lowest
+    FROM SalesByDate
+)
+SELECT
+    market_date,
+    total_sales,
+    'Highest Sales Day' AS description
+FROM RankedSales
+WHERE rank_highest = 1
+UNION
+SELECT
+    market_date,
+    total_sales,
+    'Lowest Sales Day' AS description
+FROM RankedSales
+WHERE rank_lowest = 1;
 
 
 /* SECTION 3 */
@@ -89,6 +163,17 @@ Think a bit about the row counts: how many distinct vendors, product names are t
 How many customers are there (y). 
 Before your final group by you should have the product of those two queries (x*y).  */
 
+WITH vendor_customer_cross AS (
+    SELECT vendor.vendor_id, vendor.vendor_name, customer.customer_id
+    FROM vendor    CROSS JOIN customer
+),vendor_products AS (
+    SELECT vi.vendor_id, p.product_name, vi.original_price    FROM vendor_inventory vi
+    JOIN product p ON vi.product_id = p.product_id)
+SELECT     vc.vendor_name, 
+    vp.product_name,     SUM(5 * vp.original_price) AS total_earnings
+FROM vendor_customer_cross vcJOIN vendor_products vp ON vc.vendor_id = vp.vendor_id
+GROUP BY vc.vendor_name, vp.product_nameORDER BY vc.vendor_name, vp.product_name;
+
 
 
 -- INSERT
@@ -97,18 +182,47 @@ This table will contain only products where the `product_qty_type = 'unit'`.
 It should use all of the columns from the product table, as well as a new column for the `CURRENT_TIMESTAMP`.  
 Name the timestamp column `snapshot_timestamp`. */
 
-
+CREATE TABLE product_units AS
+SELECT 
+    *, 
+    CURRENT_TIMESTAMP AS snapshot_timestamp
+FROM product
+WHERE product_qty_type = 'unit';
 
 /*2. Using `INSERT`, add a new row to the product_units table (with an updated timestamp). 
 This can be any product you desire (e.g. add another record for Apple Pie). */
 
+INSERT INTO product_units (
+    product_id,
+    product_name,
+    product_size,
+    product_category_id,
+    product_qty_type,
+    snapshot_timestamp
+)
+VALUES (
+    999,
+    'Apple Pie',
+    'large',
+    3, 
+    'unit',
+    CURRENT_TIMESTAMP
+);
+
 
 
 -- DELETE
 /* 1. Delete the older record for the whatever product you added. 
 
 HINT: If you don't specify a WHERE clause, you are going to have a bad time.*/
-
+DELETE FROM product_units
+WHERE product_name = 'Apple Pie'
+  AND snapshot_timestamp = (
+      SELECT MIN(snapshot_timestamp)
+      FROM product_units
+      WHERE product_name = 'Apple Pie'
+  );
+  
 
 
 -- UPDATE
diff --git a/05_src/sql/farmersmarket.db-journal b/05_src/sql/farmersmarket.db-journal
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index 000000000..318b10b4b
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diff --git a/bookstore_diloreto.jpg b/bookstore_diloreto.jpg
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