Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
The two consecutive integers would be 38 and 39. Hope this helps!
Answer:
1 : 6
Step-by-step explanation:
3 : 18
3 ÷ 3 : 18 ÷ 3
1 : 18 ÷ 3
1 : 6
The first expression 3^2 · 5 · 7 and the fourth expression 3^3 · 11 show a number as a product of prime numbers in exponential form.
3^2 · 5 · 7 shows 315 as a product of prime numbers 3, 5 and 7, that is 3 · 3 · 5 · 7
In 4^2 · 5 · 7, the number 4 is composite which can still be written as product 2 · 2
For the third one 3^3 · 8, the number 8 is composite which can still be written as product 2 · 2 · 2
The fourth expression 3^3 · 11 shows 99 as a product of prime numbers 3 and 11 which is 3 · 3 · 11
Using weighted average, it is found that she should score 67% on the computer science test.
- The weighted average is given by <u>each proportion multiplied by it's score</u>.
In this problem, the proportions and scores are given by:
- Proportion of 32% = 0.32 for a score of 300.
- Proportion of x for a score of 200.
- Proportion of 46% = 0.46 for a score of 300 + 200 = 500.
Then
She should score 67% on the computer science test.
A similar problem is given at brainly.com/question/24855677
