9 out of 12 = 9/12 which can be simplified to 3/4. So the team wins 3/4 of its games. 3/4 of 64 = 64 divided by 4 = 16. 16 x 3 = 48.
Answer:
17.5% per annum
Step-by-step explanation:
<u>Given:</u>
Money invested = $20,000 at the age of 20 years.
Money expected to be $500,000 at the age of 40.
Time = 40 - 20 = 20 years
<em>Interest is compounded annually.</em>
<u>To find:</u>
Rate of growth = ?
<u>Solution:</u>
First of all, let us have a look at the formula for compound interest.
Where A is the amount after T years compounding at a rate of R% per annum. P is the principal amount.
Here, We are given:
P = $20,000
A = $500,000
T = 20 years
R = ?
Putting all the values in the formula:
![500000 = 20000 \times (1+\frac{R}{100})^{20}\\\Rightarrow \dfrac{500000}{20000} =(1+\frac{R}{100})^{20}\\\Rightarrow 25 =(1+\frac{R}{100})^{20}\\\Rightarrow \sqrt[20]{25} =1+\frac{R}{100}\\\Rightarrow 1.175 = 1+0.01R\\\Rightarrow R \approx17.5\%](https://tex.z-dn.net/?f=500000%20%3D%2020000%20%5Ctimes%20%281%2B%5Cfrac%7BR%7D%7B100%7D%29%5E%7B20%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B500000%7D%7B20000%7D%20%3D%281%2B%5Cfrac%7BR%7D%7B100%7D%29%5E%7B20%7D%5C%5C%5CRightarrow%2025%20%3D%281%2B%5Cfrac%7BR%7D%7B100%7D%29%5E%7B20%7D%5C%5C%5CRightarrow%20%5Csqrt%5B20%5D%7B25%7D%20%3D1%2B%5Cfrac%7BR%7D%7B100%7D%5C%5C%5CRightarrow%201.175%20%3D%201%2B0.01R%5C%5C%5CRightarrow%20R%20%5Capprox17.5%5C%25)
So, the correct answer is <em>17.5% </em>per annum and compounding annually.
The answer is C.
x=2
y=-30
(2;-30)
Yes for the first and last the middle one should also be x^2+4
Only two real numbers satisfy x² = 23, so A is the set {-√23, √23}. B is the set of all non-negative real numbers. Then you can write the intersection in various ways, like
(i) A ∩ B = {√23} = {x ∈ R | x = √23} = {x ∈ R | x² = 23 and x > 0}
√23 is positive and so is already contained in B, so the union with A adds -√23 to the set B. Then
(ii) A U B = {-√23} U B = {x ∈ R | (x² = 23 and x < 0) or x ≥ 0}
A - B is the complement of B in A; that is, all elements of A not belonging to B. This means we remove √23 from A, so that
(iii) A - B = {-√23} = {x ∈ R | x² = 23 and x < 0}
I'm not entirely sure what you mean by "for µ = R" - possibly µ is used to mean "universal set"? If so, then
(iv.a) Aᶜ = {x ∈ R | x² ≠ 23} and Bᶜ = {x ∈ R | x < 0}.
N is a subset of B, so
(iv.b) N - B = N = {1, 2, 3, ...}
