adebayo18015:


Thank you sir for interest in helping solve this problem. I really appreciate. More knowledge and grace to you sir.

Amen boss

RoyalHighness7:
Please im interested whatsapp number 08121368722

sorry boss
it is SOLD!!!

adebayo18015:
1. A bank received 3 bad cheques per day what is the probability that in a particular day the bank will receive (I)no bad cheque (II)at least 2 bad cheques (iii) at most 1 bad cheque.

2. In a manufacturing company,for every 15 bulbs produced 3 were found defective.if 5 bulbs were produced,find the probability that (I)none is defective (II)at most 4 are defective (iii) at least 4 are defective

3. Two coins were tossed once,what is the probability that
(I) exactly two tails appear?
(II) at least two tails appear?
(III) at most three tails appear?

4. The marks scored by 1000 candidates in a test are normally distributed with a mean mark of 50 and a variance of 16,find the number of candidates whose marks are
(I) below 25
(II) above 45
(III) between 25 and 45

5.In a sta 203 exam 60% of the student passed,if a random num of 3 students were selected for investigation.what is the probability that
1.no student passed
2. Atleast one student passed
3. At most one student passed
4. No student failed
5. Atleast one student failed
6. Atmost no student failed

(5)
60% passed
p(passed)=0.6
p(failed)=0.4
q=passed
p=failed
probability is given by successive terms of (q+p)³
=q³ +3q²p +3qp² +p³
none failed(all passed)=q³=0.6³=0.216
one failed(two passed)=3q²p=3(0.6²)(0.4)=0.432
two failed (one passed)=3qp²=3(0.6)(0.4²)=0.288
three failed (none passed)=p³=0.4³=0.064
in probability,
at least = greater than or equal to
at most= less than or equal to
(i)none passed (all failed)
=p³=0.4³=0.064
(ii) at least one passed equals probability of 1,2,3 passed= 0.288 + 0.432 + 0.216=0.936
(iii)at most one passed
=p(0 passed) + p(one passed)
=0.064+0.288=0.352
(iv) at least one failed
=p(1 failed) + p(2 failed) + p( 3 failed)
=0.432 + 0.288 + 0.064
=0.784
(v)at most none failed
=q³=0.216
will drop the remaining solution if I'm chanced tonight

adebayo18015:
1. A bank received 3 bad cheques per day what is the probability that in a particular day the bank will receive (I)no bad cheque (II)at least 2 bad cheques (iii) at most 1 bad cheque.

2. In a manufacturing company,for every 15 bulbs produced 3 were found defective.if 5 bulbs were produced,find the probability that (I)none is defective (II)at most 4 are defective (iii) at least 4 are defective

3. Two coins were tossed once,what is the probability that
(I) exactly two tails appear?
(II) at least two tails appear?
(III) at most three tails appear?

4. The marks scored by 1000 candidates in a test are normally distributed with a mean mark of 50 and a variance of 16,find the number of candidates whose marks are
(I) below 25
(II) above 45
(III) between 25 and 45

5.In a sta 203 exam 60% of the student passed,if a random num of 3 students were selected for investigation.what is the probability that
1.no student passed
2. Atleast one student passed
3. At most one student passed
4. No student failed
5. Atleast one student failed
6. Atmost no student failed

(3)
When we toss two coins simultaneously then the
possible of outcomes are: (two heads) or (one
head and one tail) or (two tails) i.e., in short (H,
H) or (H, T) or (T, T) respectively; where H is
denoted for head and T is denoted for tail.
(i)
the outcomes are (H,H),(H,T),(T,T),(T,H)(total number of outcome is 4)
probability of getting exactly two tails =1/4
(ii)
atleast 2 tails
at most 2 tails
I think there's a mistake in the question it should be 3 coins

adebayo18015:
1. A bank received 3 bad cheques per day what is the probability that in a particular day the bank will receive (I)no bad cheque (II)at least 2 bad cheques (iii) at most 1 bad cheque.

2. In a manufacturing company,for every 15 bulbs produced 3 were found defective.if 5 bulbs were produced,find the probability that (I)none is defective (II)at most 4 are defective (iii) at least 4 are defective

3. Two coins were tossed once,what is the probability that
(I) exactly two tails appear?
(II) at least two tails appear?
(III) at most three tails appear?

4. The marks scored by 1000 candidates in a test are normally distributed with a mean mark of 50 and a variance of 16,find the number of candidates whose marks are
(I) below 25
(II) above 45
(III) between 25 and 45

5.In a sta 203 exam 60% of the student passed,if a random num of 3 students were selected for investigation.what is the probability that
1.no student passed
2. Atleast one student passed
3. At most one student passed
4. No student failed
5. Atleast one student failed
6. Atmost no student failed

omoh I no sure say I go get time to solve all,
(2)
15 produced,3 defective
probability of defective bulbs=0.2
probability of good bulb=1-(0.2)=0.8
p=0.2
q=0.8
the probability of there being a defective bulb is 0,1,2...5 is given by successive terms of (q +p)^5
In a sample of 5,
the probability of no defective bulb=
q^5=(0.^5=0.32768
(ii)
at most 4= probability of 0,1,2,3,4
probability of zero defective=q^5=0.32768
probability of one defective =5q⁴p
=5(0.8⁴)(0.2)
=0.4096
two defective=10q³p²
=10(0 .8³)(0 2²)
=0.2048
three=10q²p³
=10(0.8²)(0.2³)
=0.0512
four=5qp⁴
=5(0.(0.2⁴)
=0 0064
at most 4
=0+1+2+3+4(defective bulbs)
=0.32768+0.4096+0.2048+0.0512+0.0064
=0.99968
(iii)at least 4
=sum of probability of having 1,2,3,4,5 defective bulbs
=5q^5p + 10q³p² + 10q²p³+5qp⁴+p^5
already solved all except p^5 which is equal to (0.2)^5
=0.00032
therefore, probability of at least 4 defective bulbs equals the sum of having 1...5 defective bulbs
=0.4096 + 0.2048 + 0.0512 + 0.0064
+ 0.00032
=0.67242

mathefaro:
You copied the laTEX code. This would be the output:
Given that f(x)=2x-4 and g(x)=x² + 3 then f(g(x)) is
I believe you can carry on from here.

lemme carry on boss put g inside f to find f(g(x)) f(x)=2x-4 g(x)=x²+3 f(g(x)) =2(x²+3)-4 f(g(x))=2x²+6-4 =2x²+2 it can be reversed to get g(f(x))

banky200:
I need a genius to solve the logarithmic question number 20
Using the logarithmic table

genius?
for this simple looking logarithmic questions!

Ebube043:
thanks boss for the detailed explanation,pls sit how do i become this proficient,thanks as i await your favourable response

just like my mentors do tell me
you have to develop interest in the subject
and most importantly,you need to PRACTICE RELENTLESSLY.

Craggybogg:

I appreciate Bro. Buh na that second one mad pass!

x=4.7 y=0.6 will drop the full solution before 7:30

Ebube043:
If A={zER: e^x=0} and B={yER:a^y<0} determine n(AnB)

A is the empty set {}. e^x is positive for all x.
If a >0, a^y>0, because a positive number to any
power is a positive number. Therefore B is the
empty set {}. The intersection of A ({}) and B ({}) is
empty ({}).
If a=0, a^y=0 for all y other than 0 (0^0 us
undefined). Therefore, B is empty ({}). The
intersection of A({}) and B({}) is empty ({}).
If a<0, a^y
=0 for y=0,
has a positive value for y=even integers,
a negative value for y=odd integers,
an imaginary value for y=rational with an even
denominator,
and weirder values (or none) for other y values.
So B={a^1, a^3, a^5, a^7, a^9, ..., a^-1, a^-3, ...}.
The intersection of A ({}) and B({...}) is empty ({}

Ebube043:
pls sir, am a 100 lvl student studying mechanical engineering, pls am having issues with set theory that has to do with words, i need you to just guide me, so i can be able to discern any question concerning set theory on my own

Sir?I am sure you are older than I am.Though I'm still an aspirant,but i think I can help with sets.

simiife:



Thanks bro

the answers to the elevation and word problem questions is already posted on the mathematics thread
will solve the remaining 2 later

simiife:
Anyone please

Let the numbers be a and b a+b =2p.....(i) a-b =p......(ii) eliminating 2a = 3p a= 3p/2

simiife:
Anyone please

can't sketch the diagram here 40/x=tan30 tan30°=1/√3 40/x=1/√3 x=40√3

Let the numbers be a and b= a+b =2p.....(i) a-b = p......(ii) eliminating 2a = 3p a= 3p/2

Entrepreneurrr:
Help me solve this please

second question log(2)16=4x-1 log(2)2⁴=4x-1 4log(2)2=4x-1 log(2)2=1 4=4x-1 x=5/4

Entrepreneurrr:
Help me solve this please

the first question log(4)3 *log(3)4+log(2)32 converting the bases log(4)3=log(2)3/log(2)4 =log(2)3/2log(2)2 log to it's base is 1 so we have log(4)3= log(2)3/2 converting log(3)4 log(2)4/log(2)3 =2log(2)2/log(2)3 =2/log(2)3 lastly log(2)32=5log(2)2=5 computing we have (log(2)3/2)*(2/log(2)3)=1 so therefore log(4)3*log(3)4+log(2)32 =5+1=6

[quote author=simiife post=95150125][/quote] the network is bad(Airtel and MTN sucks) length of Arc @/360 * πd 60/360 * 22/7 * 42 length =22cm

[quote author=simiife post=95150125][/quote] the network is bad(Airtel and MTN sucks) length of Arc @/360 * πd 60/360 * 22/7 * 42 length =22cm

[quote author=simiife post=95150125][/quote] let his monthly income be x 20% of x=0.2x the youngest gets 25% of his income that's 0.2x * 0.25=3000 x=3000/(0.2*0.25) x=#60000

Sketch a triangle and
place M at the upper
vertex and N , O at the
two vertexes of the base.
Now, if the bisector at
angle M,meets line NO
at P( I.e,it divides the
line NO in two parts.
That is, |NP| is not
equal to|PO | because the
triangle is not an
issoceles triangle.
Bisection theorem states that;
MN/MO = NP/PO
Let us assume that,;
NP=x and PO=y
from the question ,MN=6,
MO=4 and NO=12
substitute d values;
6/4=x/y ......
crossmultiplying and
solving for y(because we were
asked to find the value of
NP,which we assumed it
to be, x), u get
y=4x/6.....equation 1
From the triangle, NP+PO=NO ( x
+y=12)
then,
x+y=12.......equation2
substitute 4x/6,for y in
equation 2.
x+y=12
x+4x/6 = 12
multiply both sides by 6
6x+4x=72
10x=72
divide both sides by 10
Hence, x= NP=7.2

simiife:

Good evening bro, been busy, would have posted since

do justice to the simple interest question using the formula S.I=(PRT)/100 where S.I=interest P=principal R=rate T=time

simiife:

Good evening bro, been busy, would have posted since

( (a²-a)C2=(a²-a)C4 let (a²-a)=x xC2=xC4 comparing both sides of the equation, 2 is not equal to 4 so definitely, x=2+4 x=6 (a²-a)=x a²-a=6 a²-a-6=0 a+2a-3a-6=0 a(a+2)-3(a+2)=0 (a-3)(a+2)=0 a=3 or -2 therefore a=3 there is no diagram in the third question

simiife:

Good evening bro, been busy, would have posted since

2x-y=4 x+y=2 from x+y=2,x=2-y substituting into 2x-y=4 2(2-y)-y=4 4-4y-y=4 5y=0 y=0 x=2-y x=2 points x,y=(2,0) and (4,3) formula for distance (√((x2-x1)²+(y2-y1)²) x1=2, x2=4 y1=0 y2=3 distance=(√(4-2)²+(3-0)²) =√((2)²+(3)1) =√13

some months ago,I saw a post on this platform about world scholarship initiative.A link was provided and I registered,the exam was meant to be a cbt,but due to the pandemic enforced lockdown it was done online.I just received a SMS now inviting me to the stakeholders forum and award presentation.But I was told by a friend that after the exam,there will be a follow up interview.
Please I need more information from those who have gone for the interview before how the interview is gonna be.
thanks!

Sweetcrude19:
OK but my advice for u is to channel ur effort towards securing admission into one of the following schools u listed above. Because mathematics for OAU might be quite different from Unilorin. Thanks

Thanks boss,I really appreciate

yusman14:
I guess you school in OAU...you can only get pq for general courses like 101,102,201,202 on campus around #800-1k...but you can get others from your seniors in form of question papers because they are departmental courses...they are not available for sale

thanks boss

Sweetcrude19:
Which school in particular do u need it's past question

unilorin or Oau boss

Good day nairaland family,I have been practicing some engineering mathematics topics for a while now.After each topic,I always try to solve the solve the exercises at the end of course each topic.but my problem now, is that I have been sourcing for past questions online to solve but I couldn't get any.
I need past questions on the following courses
Math 101-Elementary Mathematics 1
Math 102-Elementary mathematics 2
Math 103-Elementary mathematics 3
Math 201-Introduction to mathematical methods
Math 202-Elementary differential equations
Math 203-sets,Logic and algebra
math 204-Linear algebra 1
math 205-linear algebra 2
I haven't actually come covered Math209(intro. to Numerical analysis) and I haven't touched math207(Real analysis)
I will be glad if anyone could help me with all or some of the past questions.God bless you all!
fynestboi
Richiez
olawalebabs

I need a clean /mint phone 3gb or 4gb ram battery(4000-5000mah) Budget 20-24k

buchichris:
A week old REDMI 9A. 37k. Receipt and carton stil dey Call/ watsap 07035024341

boss abeg I get 25k

ojonugba2014:

One love

come what app make we talk better zero nine zero seven three six nine nine three three four