Perhaps the easiest way to solve this problem is to convert 13 into a fraction that has the same denominator as 7 5/8.
Convert both to improper fractions:
7 5/8 turns into 61/8, and
13 turns into 104/8.
Then, subtract 61/8 from 104/8:
104/8-61/8=43/8.
Simplify (mixed fraction):
5 3/8.
The second ribbon has a length of 5 3/8 meters.
In statistics, the amount of degrees of freedom is
the quantity of values in the final computation of a statistic that are free to
differ. In this case, you can get the answer by adding the number of bags
and subtracting 1.
So in computation, this would look like: 3 + 2 + 1 + 2 - 1 =
7
Therefore, 7 is the degrees of freedom.
First we need to identify if the data is qualitative or quantitative.
The data is average number of people living in the homes.
Qualitative data as its name indicates is an attribute or characteristic. It can not be measured e.g color. Quantitative data is such a data which can be counted or measured.
Since the average number of people can be counted and measured, the data is Quantitative.
In an observational study the individuals are observed. In the given case, Kira did not observed the individuals to gather the data, rather she used an Online resource to gather the data.
Therefore, the correct answer will be:
Kira used published data that is quantitative.
£87.00
Emily's dad pays 3 parts of the meal
Divide £52.20 by 3 to find one part of the ratio
= £17.40 ← 1 part of the ratio
2 parts = 2 × £17.40 = £34.80 ← Emily's share
total cost = £52.20 + £34.80 = £87.00
<span>Let a_0 = 100, the first payment. Every subsequent payment is the prior payment, times 1.1. In order to represent that, let a_n be the term in question. The term before it is a_n-1. So a_n = 1.1 * a_n-1. This means that a_19 = 1.1*a_18, a_18 = 1.1*a_17, etc. To find the sum of your first 20 payments, this sum is equal to a_0+a_1+a_2+...+a_19. a_1 = 1.1*a_0, so a_2 = 1.1*(1.1*a_0) = (1.1)^2 * a_0, a_3 = 1.1*a_2 = (1.1)^3*a_3, and so on. So the sum can be reduced to S = a_0 * (1+ 1.1 + 1.1^2 + 1.1^3 + ... + 1.1^19) which is approximately $5727.50</span>
