Can I get some help with questions? (2 Viewers)

[CM_Tutor]

this got serious real quick lol

Yes, you can blame me for that.

 

[jimmysmith560]

Would this help?

Regarding this approach, the following formula is being used:

Alternatively, I suggest you refer to the following thread, where this specific question was covered in depth:

Is this part of our syllabus?

boredofstudies.org

I hope this helps!

 

[5uckerberg]

Wait so how does showing that m=n show their independent events?
View attachment 33612

In the Venn diagram, the two circles are separated from one another and as such the union of the two cases is one probability added onto the other. Next, for mutually inclusive the idea is this inclusion-exclusion principle. That is clear,
Now, for our independent events what is the story there so basically what happens is that the events have no connection with one another so pretty much what happens is that if you want the two events together you multiply them. It is like oh a boy will be born from the mother's womb and it is also sunny out there. To be fair one can envision this as the two circles being separated situation. The way of attack after a bit of thought is this Independent or mutually exclusive circles in the Venn Diagram separate the circles, and mutually inclusive and dependent the circles in the Venn Diagram are together.

Think of it like this being excluded from a group is the harsh term for being independent of the group and included is when people are together and such. Just think of you and your friend folding arms looking in the opposite direction as mutually exclusive.

 

[5uckerberg]

Wait so how does showing that m=n show their independent events?
View attachment 33612

 

[CM_Tutor]

@CM_Tutor

Have you looked at my post here where I provided a different approach to this question?

 

[5uckerberg]

@5uckerberg

Yep, Cm_tutor's method explicitly shows what is going on but what I did is my interpretation. I sometimes feel bad not knowing but it is good to be able to learn a few things ourselves.

 

[cossine]

View attachment 35220

So you would end up with geometric series = num_minutes_in_a_year, where n= num_times_finger_snapped. There is no way to solve this numerically so you will need to use inspection

 

[jimmysmith560]

The official solution is as follows:

This question also requires a knowledge of logarithms. Don’t stress if you don’t understand it yet!

To solve a problem like this, it is often helpful to draw a diagram and/or table to illustrate what is happening. It then becomes easier to spot a pattern. We will draw a table to compare the number of finger snaps (n) with the total time elapsed in minutes (t):

​

Can we find some mathematical expression that will give us the time elapsed for a certain number of finger snaps? Since the time elapsed between finger snaps always doubles, we would expect powers of 2 to be involved in the expression. In fact, we notice that the numbers in the time elapsed column are all 1 less than a power of 2. We could try the expression

​

but this doesn’t quite work. With a little bit of thought, you should realise that

​

is the correct expression. Since we can now calculate the time elapsed for any given number of finger snaps, we can work out how many finger snaps fit into a year. We do this by first finding the number of minutes in a year.

We could of course look this up on the internet; but we really should be able to work it out for ourselves! One year is 365 days, which is 365 × 24 hours, i.e. 8760 hours. This in turn is 8760 × 60 minutes; so there are 525600 minutes in a year.

Now we ask the question: for which value of n (for how many finger snaps) do we get a value of t that is close to the number of minutes in a year? In other words, we want to solve , which gives

​

Since we can only have a whole number of finger snaps, we have shown that by following the pattern, we would only snap our fingers 20 times in one year.

I hope this helps!

 

[brent012]

This one is a similar problem: https://www.gocomics.com/foxtrot/2006/09/10

 

[jimmysmith560]

how do you get the annual percentage change

In all cases, for the percentage change, we use the formula:

​

a) Percentage increase = . Hence the average annual percentage income increase over the 6 years is .

b) For men, percentage increase = . Hence the average annual percentage increase over the 15 years is .

For women, percentage increase = . So the average annual percentage increase over the 15 years is .

c) Percentage increase = . Hence the average annual percentage increase over the 30 years is .

d) In shops, percentage increase = . So the average annual percentage increase over the 3 years is .

In pubs, percentage change = . So the average annual percentage decrease over the 3 years is .

I hope this helps!

 

[cossine]

View attachment 35247



Seems simple enough right? The answer the university gave is 1/3 + 1/2 = 5/6

But i wrote an equation for it and got a different answer being 3/4
View attachment 35248

To explain the equation.
x is the amount of lemonade in the jug to start off with.
1/3 of x was sold and then only one half of the jug was left

Am i doing something wrong? or is the question wrong?

So x is the current amount of lemonade as a fraction which may be less than one. You are taking 1/3 of current amount of lemonade not the full jug.

 

[jimmysmith560]

Pretty much cossine's advice, solving this question algebraically would be as follows:





Solving the equation will allow us to find the required fraction:



,

I hope this helps!

 

[yanujw]

Put more simply;

Any lemonade at the start of the day has either been sold or remains in the jug. Now the question says 1/3 of a jug (no particular relation to the initial amount) was sold, and 1/2 of a jug remains. So the lemonade at the start is 1/3 + 1/2 = 5/6. Interpret the values as proprotions of a jug, not relations to the initial or final amounts.

 

[jimmysmith560]

Your assuming that x = 1. That's the only way you know that 1/3 of x is really 1/3

Eg. If the jug was 5/6 then one-third of the jug would be 5/18 not 1/3

In your example if the jug was 1 then 1/3 of 1 is 1/3. Hence you can do the subtraction. But we don't know that which is why we go
x (being the unknown amount in the jug) - (1/3 * x) being 1/3 of the unknown amount. View attachment 35251

In addition to the above, and in order to confirm the original answer, there is a slight variant of this question from the University of Wollongong, where the only difference is that the numbers were swapped (which leads to the same answer), as follows:

Katie and Tyler are working at their lemonade stand. They have sold of a pitcher of lemonade and now have of a pitcher left to sell. What fraction of a pitcher of lemonade did they start with?

The solution to the above version of the question is as follows:

To find out how much lemonade they started with, we can add how much they sold to how much they have left:



Therefore, Katie and Tyler started with of a pitcher of lemonade.

 

[Eagle Mum]

Perhaps we could reword the question as:
Katie and Tyler are eating pizza. They have eaten

a pizza and now have

of a pizza left. What fraction of a pizza did they start with?

 

[5uckerberg]

Wait @Hivaclibtibcharkwa which uni course is this and which uni is this from?

 

[5uckerberg]

It has to be UWS. Right?

 

[jimmysmith560]

View attachment 35475

First, perform the necessary conversions:

• To go from g to kg, we divide the mass value by 1000. This means that 8.9g is 0.0089kg
• To go from to , we multiply the volume value by 1000. This means that

Now that you have the values required, use .

.

In scientific notation, this would be expressed as , matching the answer provided.

I hope this helps!

 

[cossine]

Can someone help me with these question, like with question b would the null hypothesis be that it’s less than 70 or would the alternative hypothesis say that?

The null hypothesis is always an equality. The reason for this is computation.

Because the definition p-value is

p-value = P(getting test statistic that is more contradictory or equally contradictory to the null hypothesis | H0 )

It may help to read Probability and Statistics for Engineering Sciences by Jay L. Devore

It will explain things in more detail

 

[cossine]

Ah sorry just realised I didn’t put in the imahe

So give the question a go first and I will check the working out.

In this case

H0: mean_weight = 70
Ha: mean_weight <= 70.