Answer:
11 of 20p, 22 of 10p and 33 of 5p
Step-by-step explanation:
Eva has 20p, 10p and 5p coins, total of £6.05 = 605p
Let 20p=x, 10p=y, 5p=z
<u>Then</u>:
- 20x + 10y + 5z = 605
- y : x = 2 : 1 ⇒ x= y/2
- y : z = 2 : 3 ⇒ z= 3y/2
<u>Rewriting the first equation considering next two:</u>
- 10y + 10y + 7.5y = 605
- 27.5y = 605
- y= 605/27.5
- y= 22
- x= y/2 = 22/2 = 11
- z = 3y/2 = 3*11 = 33
<u>Answer:</u> 11 of 20p coins, 22 of 10p coins and 33 of 5p coins
Answer:
Step-by-step explanation:
common difference d=3-1=2
first term a=1
an=a+(n-1)d
2n-1=1+(l-1)2
2n-1=1+2l-2
2n-1=2l-1
l=n
(i used l for number of terms)
number of terms=n

Answer:
A sample size of 6755 or higher would be appropriate.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the zscore that has a pvalue of
.
The margin of error M is given by:

90% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
52% of Independents in the sample opposed the public option.
This means that 
If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?
Sample size of size n or higher when
. So







A sample size of 6755 or higher would be appropriate.
It will take 3.35........