NisforNicky:


Pls help with this.......


I need the workings

x + y/2 = 3.2
x + y/4 = 2.3

Solve simultaneously y/4 = 0.9
y = 3.6kg

Mass of empty bottle x = 3.2 -1.8 = 1.4 kg

ThankGod001:
Yes, its online so u can have ur cal beside you

I doubt o

A call was made to dragnet and it was stated that if Calcs were needed,it would be made available in the system.

femi4:
x + y/2 = 3.2 x + y/4 = 2.3
Solve simultaneously y/4 = 0.9 y = 3.6kg
Mass of empty bottle x = 3.2 -1.8 = 1.4 kg

wow dat was fast thanks bro
pls don't forget the rest

Second moving average of order 3

(50+10+70)/3 = 130/3 = 43.3

cc femi4

now, this one below is from the recently concluded SPDC

mete3

femi4:
Second moving average of order 3

(50+10+70)/3 = 130/3 = 43.3

does this mean that we go to the second value on the data log, and take three values and then find their average


dats the only thing I understand about dis

Are this question also for Snepco

NisforNicky:
cc femi4

mete3:
Are this question also for Snepco

spdc but they ask similar questions

all the ones above are SNEPCo tho

NisforNicky:
cc femi4

1/x + 1/y = 1/4

1/x + 1/6 = 1/4

X = 12days

NisforNicky:


does this mean that we go to the second value on the data log, and take three values and then find their average


dats the only thing I understand about dis

Exactly

femi4:
1/x + 1/y = 1/4
1/x + 1/6 = 1/4
X = 12days

no 12 in options

femi4:
Exactly

thanks a lot
this helped too https://www.statisticshowto.com/moving-average/

NisforNicky:


no 12 in options

Well , I was surprised too

NisforNicky:

spdc but they ask similar questions
all the ones above are SNEPCo tho

Okay.. Thanks
Buh pls, Did u hav them in pdf or what.

mete3:


Okay.. Thanks

Buh pls,
Did u hav them in pdf or what.

yh
but too big to upload on Nairaland

ill try posting them as screenshots later when i buy data

using free mb now

I hv posted virtually all the math sef

jxt varbal left

The set doesn't have a mode ( no repetition of numbers)

NisforNicky:

yh but too big to upload on Nairaland
ill try posting them as screenshots later when i buy data
using free mb now
I hv posted virtually all the math sef
jxt varbal left

Can i pm u Boss, I need it pls.

femi4:
The set doesn't have a mode ( no repetition of numbers)

exactly

u think these are negative questions?

mete3:

Can i pm u Boss, I need it pls.

no need pls
il post it here later for everyone

NisforNicky:


no need pls

il post it here later for everyone

I already did but no p.
the thing is, am nt using android for nw nd was unable to expand the size of qtns been posted.

reason why i ask u to help me snd them in pd format.

femi4:
Well , I was surprised too

I don't think they re working at d sam rate

NisforNicky:


exactly

u think these are negative questions?

May be

NisforNicky:


Even more

In a game of cards, there are 4 kings.
Therefore, prob that the cards are kings will be

4/52 x 3/51 = 1/221

As long as it is not stated otherwise in the question, drawing should always be without replacement.

femi4:
1/x + 1/y = 1/4

1/x + 1/6 = 1/4

X = 12days

Lol...Thank God I'm not the only one..
I started doubting the little knowledge I have when I could not find 12 in the options.

ivumar:


In a game of cards, there are 4 kings.
Therefore, prob that the cards are kings will be

4/52 x 3/51 = 1/221

As long as it is not stated otherwise in the question, drawing should always be without replacement.

where did you get 3/51 pls

NisforNicky:



where did you get 3/51 pls

Based on the without replacement I emphasized on.

If initially, there were 4 kings. How many kings will remain after you have drawn one out?
Also the initial number of cards was 52. How many will remain after one card is drawn out?

You get right?

ivumar:


Based on the without replacement I emphasized on.

If initially, there were 4 kings. How many kings will remain after you have drawn one out?
Also the initial number of cards was 52. How many will remain after one card is drawn out?

You get right?

no.

weren't the cards drawn simultaneously?

NisforNicky:


no.

weren't the cards drawn simultaneously?

simultaneously was not specified..
Even if it was, it does not matter. You still have to treat the cards individually.

Reason being that, if you are to dig through layers of abstraction, then two objects picked 'simultaneously' will still differ in the time picked even if the difference in time is usually so small(in the order of micro seconds). Therefore, there will always be a first card picked and a second(although they were picked simultaneously). That's why we account for the first card picked, after which we account for the second, bearing in mind, that a card has been picked already.

ivumar:


simultaneously was not specified..
Even if it was, it does not matter. You still have to treat the cards individually.

Reason being that, if you are to dig through layers of abstraction, then two objects picked 'simultaneously' will still differ in the time picked even if the difference in time is usually so small(in the order of micro seconds). Therefore, there will always be a first card picked and a second(although they were picked simultaneously). That's why we account for the first card picked, after which we account for the second, bearing in mind, that a card has been picked already.

oh thanks

I finally get it

you are a good tutor bro



pls help attempt the rest

NisforNicky:


More

From the sequence 3, 4, 1, -3, -4, -1....., it can be observed that the nth term is gotten from subtracting the (n-2)th term from the (n-1)th term.
In other words,
nth term = (n-1)th term - (n-2)th term. Okay lets validate..
According to the sequence, the 3rd term is 1. Lets check if it tallies with the emboldened.
3rd term = 2nd term - 1st term.
i.e, 3rd term = 4 - 3 = 1.
This shows correspondence. Now you can check for 4th, 5th and 6th term for yourself and see that they agree with the emboldened.

Solving for the 7th term, we have
7th = -1 - (-4) = -1 + 4 = 3
Therefore, the 7th term is 3.

Now we solve for the difference between btw the 7th and 6th term, which is
3 - (-1) = 3 + 1 = 4.
Answer not in the options
I guess they mistakenly did '3 - 1' instead

NisforNicky:


More

The question on angle of elevation is, I think incorrect. Or what do you guys think... because I don't get how angle of elevation can be greater than 90 degrees.

NisforNicky:


More

The graph of y=x and y=x² intersects when, at the same value of x, both graphs have the same value of y.
Observing the options, there are only two possible values... 0 and 1. So lets use them to test.
for the graph, y = x, when x = 0, y = 0.
Will it be same for graph of y = x²? lets check.
when x = 0, y = 0², therefore, y = 0.
Therefore, for both graphs, when x=0, y will also = 0. Therefore, they intersect at (0,0)

Next, test for when x = 1, you will also discover that for both graphs, when x = 1, y will also = 1. Therefore, they also intersect at (1,1).




Alternatively, you can solve using the equation of both graphs.
y1 = x
y2 = x².
Recall that at the point of intersection, y1 must = y2
So at what value of x can y1 = y2
We solve this by equating y1 to y2, i.e, y1 = y2, i.e x = x². Lets solve
x = x²
x² - x = 0
Factorize
x(x - 1) = 0
Therefore, x = 0 or x = 1
Input this values of x in each of the equations to get y.
When x is 0, you will get 0 as y for both graphs
And when x is 1, you will get 1 for both graphs.
Therefore point of intersection is (0,0) and (1,1)