In a certain country the life expectancy for women in 1990 was 45 and in 2000 it was 85?years. Assuming that life expectancy between 2000 and 2100 increases by the same percentage as it did between 1900 and 2000,what will life expectancy be for women in 2100? Assuming the life expectancy between 2000 and 2100 will increase by the same percentage as it did between 1900 and 2000, the life expectancy for women will be —— years

49145 years

Step-by-step explanation:

In a certain country the life expectancy for women in 1990 was 45 and in 2000 it was 85?years.

Assuming that life expectancy between 2000 and 2100 increases by the same percentage as it did between 1900 and 2000,what will life expectancy be for women in 2100?

In 10 years, the expectancy increased by 85/45 = 17/9

between 2000 and 2100, it will increas by 10 time 10 years, so expected expectancy is [tex]85*(\frac{85}{45})^{10} = 49145[/tex] years

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[ (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

[tex]Can somebody solve some basic algebra[/tex]