Lamont’s grandmother started a stamp collection in 1914 (year 1) and added the same number of stamps to the collection each year. when lamont’s mother received the collection in 1954, there were 203 stamps in the collection. lamont’s mother continued to add the same number of stamps to the collection each year. when lamont received the collection in 1990, there were 383 stamps in the collection. if lamont continued to add the same number of stamps each year, how many stamps were in the collection at the end of 2000 (year 86)?
a. 204
b. 293
c. 433
d. 663

Option C is the correct answer.

Step-by-step explanation:

In 1954, there were 203 stamps in the collection and in 1990 there were 383 stamps in the collection.

We can find number of stamps collected per year using this

Stamps collected from 1954 to 1990 = 383 - 203 = 180

Number of years = 1990 - 1954 = 36

Stamps collected per year

This is an example of arithmetic progression given like

      a, a+5, a +10.........

Now we need to find first term of this AP

We have term of an AP given by a + (n-1)d

We also have n 1954, there were 203 stamps in the collection ( n = 1954-1914 + 1 = 41)

      a + (n-1)d = 203

      a + (41 - 1) x 5 = 203

      a = 3

Now we need to find number of stamps were in the collection at the end of 2000 (n = 2000 - 1914 +1 = 87),

      Number =  a + (n-1)d = 3 + (87-1)5 = 433

Number of stamps were in the collection at the end of 2000 = 433

Option C is the correct answer.

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )

step-by-step explanation: 4 cause 2+2=4