Answer:
<h2>
£1,330.46</h2>
Step-by-step explanation:
Using the compound interest formula 
A = amount compounded after n years
P = principal (amount invested)
r = rate (in %)
t = time (in years)
n = time used to compound the money
Given P = £1200., r = 3.5%, t = 3years, n = 1 year(compounded annually)
Value of Charlie's investment after 3 years is £1,330.46
Answer:
-23x^3+20x^4+25x^2+84x-84
Step-by-step explanation:
1 Expand by distributing sum groups.
4x^2(3x+5x^2-6)-7x(3x+5x^2-6)+14(3x+5x^2-6)
2 Expand by distributing terms.
12x^3+20x^4-24x^2-7x(3x+5x^2-6)+14(3x+5x^2-6)
3 Expand by distributing terms.
12x^3+20x^4-24x^2-(21x^2+35x^3-42x)+14(3x+5x^2-6)
4 Expand by distributing terms.
12x^3+20x^4-24x^2-(21x^2+35x^3-42x)+42x+70x^2-84
5 Remove parentheses.
12x^3+20x^4-24x^2-21x^2-35x^3+42x+42x+70x^2-84
6 Collect like terms.
(12x^3-35x^3)+20x^4+(-24x^2-21x^2+70x^2)+(42x+42x)-84
7 Simplify.
-23x^3+20x^4+25x^2+84x-84
<h2>Verbal Phrase to Mathematical Phrase</h2><h3>
Answer:</h3>
<h3>
Step-by-step explanation:</h3>
Let that number be 
.
The quotient of a number and 
is 
.
"Increased" is just another term for "plus (+)" so 
increased by The quotient of a number and is 
And it tells us that it is 
so it must be
Answer:
I think it is $156
Step-by-step explanation:
Multiply the numbers
Answer:
We accept H₀ , we do not have enought evidence for rejecting H₀
Step-by-step explanation:
Normal Distribution
sample size n = 60
standard deviation σ = 15
1.Hypothesis Test : Is a one tailed-test on the right
H₀ null hypothesis μ₀ = 50
Hₐ alternative hypothesis μ₀ > 50
2.-We will do the test for a significance level α = 0,01 tht means for a 99% interval of confidence
then z(c) = 2.32
3.- We compute z(s)
z(s) = [ ( μ - μ₀ ) /( σ/√n ) ⇒ z(s) = ( 2 * √60 ) / 15
z(s) = 15.49/15 ⇒ z(s) = 1.033
4.- We compare values of z(c) and z(s)
z(s) < z(c) 1.033 < 2.32
z(s) is in the acceptance region so we accept H₀ , we do not have enough evidence for rejecting H₀
