<u><em>Answer:</em></u>
24.2%
<u><em>Explanation:</em></u>
<u>The percentage of people choosing strawberry can be calculated as follows:</u>

<u>We have:</u>
Number of people choosing strawberry = 15 people
Total number of surveyed people = 17 + 10 + 17 + 3 + 15 = 62 people
<u>Substitute in the above rule to get the percentage as follows:</u>
%
Hope this helps :)
Answer:
6
because when you subtract 6 from 6 you get 0
6 - 6 = 0
Answer: the value of her investment after 4 years is £8934.3
Step-by-step explanation:
The formula for determining compound interest is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount invested.
t represents the duration of the investment in years.
From the information given,
P = 8000
r = 2.8% = 2.8/100 = 0.028
n = 1 because it was compounded once in a year.
t = 4 years
Therefore,
A = 8000(1+0.028/1)^1 × 4
A = 8000(1+0.028)^4
A = 8000(1.028)^4
A = £8934.3 to the the nearest penny
Answer:
c = 3 c = -1/2
Step-by-step explanation:
(4c-5)^2= 49
Take the square root of each side
sqrt((4c-5)^2)= ±qrt(49)
4c-5 = ±7
Separate into two equations
4c-5 = 7 4c-5 = -7
Add 5 to each side
4c-5+5 = 7+5 4c-5+5 =-7+5
4c =12 4c = -2
Divide by 4
4c/4 = 12/4 4c/4 = -2/4
c = 3 c = -1/2
Answer:
proof below
Step-by-step explanation:
Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)
Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;
(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2
= 2(2a^2 + 2a + 2b^2 + 2b + 1)
Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.