Last year Oscar bought a packet of biscuits every week and found 30 of 3 these packets contained broken biscuits. What is the relative frequency
of this event? Answer correct to 2 decimal places. Explain why you
chose the numbers in your calculation

Relative Frequency = 0.58

Step-by-step explanation:

Given

A packet of biscuit every week

Number of packets that contains broken biscuits = 30

Required

Relative Frequency

First, the total packet of biscuit bought needs to be calculated.

The question says Oscar bought a packet every week of the year and there are 52 weeks in a year.

This means that Oscar bought 52 packets of biscuits.

So, we can conclude that 30 out 52 packets contains broken biscuits.

The relative frequency is then calculated as this;

Relative Frequency = Number of packets that contains broken biscuits divided by total number of packets.

Relative Frequency = 30/52

Relative Frequency = 0.5769230769

Relative Frequency = 0.58 (Approximated).

Hence, the relative frequency of the broken packets of biscuits is 0.58

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!