Edexcel A level Maths 2023 paper 3 UOMS

Feel free to correct this

Venn diagram

A only:0.13,A∩B:0.25,B only:0.05,B∩C: p,C only:0.3,Outside the circles:q

1(a)P(A)=0.38

(b)p=0.2, q=0.07

(c)P(A|B')=0.325

Binomial

B(40,1/7), test n=110,9

2(a)P(T=6)=0.173

P(T<3)=0.0615

(b)P(T<3 exactly twice)=0.158

(c)Reject H₀

Large data set

Sum of x=9, Sum of x²=4341

3(a)Consider tr, treat tr as zero as ≤0.05mm rounds to 0

(b)Mean=2.12, SD=4.37

(c)May to Oct only not representable

The great storm so actual mean is less

Normal distribution

N(175.4,6.8²), 52 sample

4(a)P(X>180)=0.2494

(b)Accept H₀

(c)p=0.0562

Discrete distrbution

20,50,80,100

a,b,c,d

P(S|{X=x})=k/x

P(S∩{X=x}) constant for all x

5(a)bk/50=(S∩{X=50})

ck/80=(S∩{X=80})

As P(S∩{X=20}) constant for all x, bk/50=ck/80

c=8/5b

(b)d=2b,a=2/5b,a+b+c+d=1,b=1/5

2/25,1/5,8/25,2/5

(c)k/100=0.3,k=30,P(S|{X=20})=30/20=1.5>1 so not reasonable

Histogram

N(14.9,3.2²)sth like that, y=kxe⁻ˣ for 0≤x≤4, unit of x is ten years

6(a)0.538

(b)Not symmetrical, median not the same etc

(c)Show that ∫₀ⁿxe⁻ˣdxdx=1-(n+1)e⁻ⁿ

(d)k(∫₀⁴ xe⁻ˣdx)=90 so k=99

(e)99/90∫₁³xe⁻ˣdx=0.5902

(f)Sample size=90 is too small to estimate

1D suvat

u=0, t=5, a=3.2

1(a)v=16

(b)s=40

1D forces

5kg, Pulling force=28, a=1.4

2(a)R=49

(b)Friction=21

(c)mu=0.43

Parametric

v=(t²-3t+7)i+(t²-3)j

3(a)|v₀|=7.62

(b)t=2

(c)a=(2t-3)i+4tj

(d)t=1.5

Vector suvat

u=-16i-3j, v=2.4i+j, t=5, at time T s=4i+cj

4(a)4.47

(b)T=10,c=10

Projectile

u=28, x=40, y=20

5(a)Horizontal:s=vt

40=28cosθT

T=10/(7cosθ)

(b)Vertical:s=ut+1/2at²

20=(28sinθ)10/(7cosθ)+1/2(-9.8)(10/(7cosθ))²

20=40tanθ-10(1+tanθ²)

tan²θ-4tanθ+3=0

(c)36

(d)Spin of the ball, accurate value of g etc

Ladder

tanθ=3/4

6(a)Ssinθ(2a)=Mgcosθ(a)

S=1/2Mgcotθ

(b)Vertical:R=mg

F=μR=μmg

Horizontal:F=S

μmg=1/2Mgcotθ=3/2Mg

μ=2/3

(c)Magnitude=√13/3Mg

(d)Farther away so moment increases so S is bigger

Feel free to correct this

Venn diagram

A only:0.13,A∩B:0.25,B only:0.05,B∩C: p,C only:0.3,Outside the circles:q

1(a)P(A)=0.38

(b)p=0.2, q=0.07

(c)P(A|B')=0.325

Binomial

B(40,1/7), test n=110,9

2(a)P(T=6)=0.173

P(T<3)=0.0615

(b)P(T<3 exactly twice)=0.158

(c)Reject H₀

Large data set

Sum of x=9, Sum of x²=4341

3(a)Consider tr, treat tr as zero as ≤0.05mm rounds to 0

(b)Mean=2.12, SD=4.37

(c)May to Oct only not representable

The great storm so actual mean is less

Normal distribution

N(175.4,6.8²), 52 sample

4(a)P(X>180)=0.2494

(b)Accept H₀

(c)p=0.0562

Discrete distrbution

20,50,80,100

a,b,c,d

P(S|{X=x})=k/x

P(S∩{X=x}) constant for all x

5(a)bk/50=(S∩{X=50})

ck/80=(S∩{X=80})

As P(S∩{X=20}) constant for all x, bk/50=ck/80

c=8/5b

(b)d=2b,a=2/5b,a+b+c+d=1,b=1/5

2/25,1/5,8/25,2/5

(c)k/100=0.3,k=30,P(S|{X=20})=30/20=1.5>1 so not reasonable

Histogram

N(14.9,3.2²)sth like that, y=kxe⁻ˣ for 0≤x≤4, unit of x is ten years

6(a)0.538

(b)Not symmetrical, median not the same etc

(c)Show that ∫₀ⁿxe⁻ˣdxdx=1-(n+1)e⁻ⁿ

(d)k(∫₀⁴ xe⁻ˣdx)=90 so k=99

(e)99/90∫₁³xe⁻ˣdx=0.5902

(f)Sample size=90 is too small to estimate

1D suvat

u=0, t=5, a=3.2

1(a)v=16

(b)s=40

1D forces

5kg, Pulling force=28, a=1.4

2(a)R=49

(b)Friction=21

(c)mu=0.43

Parametric

v=(t²-3t+7)i+(t²-3)j

3(a)|v₀|=7.62

(b)t=2

(c)a=(2t-3)i+4tj

(d)t=1.5

Vector suvat

u=-16i-3j, v=2.4i+j, t=5, at time T s=4i+cj

4(a)4.47

(b)T=10,c=10

Projectile

u=28, x=40, y=20

5(a)Horizontal:s=vt

40=28cosθT

T=10/(7cosθ)

(b)Vertical:s=ut+1/2at²

20=(28sinθ)10/(7cosθ)+1/2(-9.8)(10/(7cosθ))²

20=40tanθ-10(1+tanθ²)

tan²θ-4tanθ+3=0

(c)36

(d)Spin of the ball, accurate value of g etc

Ladder

tanθ=3/4

6(a)Ssinθ(2a)=Mgcosθ(a)

S=1/2Mgcotθ

(b)Vertical:R=mg

F=μR=μmg

Horizontal:F=S

μmg=1/2Mgcotθ=3/2Mg

μ=2/3

(c)Magnitude=√13/3Mg

(d)Farther away so moment increases so S is bigger

(edited 10 months ago)

Original post by Anonymo us

Edexcel A level Maths 2023 paper 3 UOMS

Feel free to correct this

Venn diagram

1(a)0.38

(b)0.2

(c)0.07

(d)0.325

Binomial

B(40,1/7), test n=110,9

2(a)0.173

P(T<3)0.0615

(b)P(T<3 exactlytwice)=0.158

(c)Accept H1

Large data set

Sum of x=9, Sum of x²=4341

3(a)Consider tr, treat tr as zero as ≤0.05mm rounds to 0

(b)2.12,4.37

(c)May to Oct only not representable

The great storm so actual value is less

Normal distribution

N(175.4,6.8²), 52 sample

4(a)0.2494

(b)Accept H0

(c)0.0562

Discrete distrbution

20,50,80,100

a,b,c,d

P(S|{X=20})=k/x

P(S∩{X=20}) constant for all x

5(a)Show thtat c=5/8b

(b)2/25,1/5,8/25,2/5

(c)k/100=0.3,k=30,P(S|{X=20})=30/20=1.5>1 so not reasonable

Histogram

N(14.9,3.2²)sth like that, y=kxe⁻ˣ for 0≤x≤4, unit of x is ten years

6(a)0.538

(b)Not symmetrical, median not the same etc

(c)Show that ∫₀ⁿxe⁻ˣdxdx=1-(n+1)e⁻ⁿ

(d)k(∫₀⁴ xe⁻ˣdx)=90 so k=99

(e)0.5902

(f)Sample size=90 is too small to estimate

1D suvat

u=0, t=5, a=3.2

1(a)v=16

(b)s=40

1D forces

5kg, Pulling force=28, a=1.4

2(a)R=49

(b)Friction=21

(c)mu=0.43

Parametric

v=(t²-3t+7)i+(t²-3)j

3(a)|v0|=7.62

(b)t=2

(c)a=(2t-3)i+4tj

(d)t=1.5

Vector suvat

u=-16i-3j, v=2.4,1, t=5, at time T s=4i+cj

4(a)-4i+2j

(b)4.47

(c)T=10,c=10

Projectile

u=28, x=40, y=20

5(a)T=10/(7cosθ)

(b)tan²θ-4tanθ+3=0

(c)36

(d)Spin of the ball, accurate value of g etc

Ladder, tanθ=3/4

6(a)S=1/2Mgcotθ

(b)2/3

(c)√13/3Mg

(d)Farther away so moment increase so S bigger

Feel free to correct this

Venn diagram

1(a)0.38

(b)0.2

(c)0.07

(d)0.325

Binomial

B(40,1/7), test n=110,9

2(a)0.173

P(T<3)0.0615

(b)P(T<3 exactlytwice)=0.158

(c)Accept H1

Large data set

Sum of x=9, Sum of x²=4341

3(a)Consider tr, treat tr as zero as ≤0.05mm rounds to 0

(b)2.12,4.37

(c)May to Oct only not representable

The great storm so actual value is less

Normal distribution

N(175.4,6.8²), 52 sample

4(a)0.2494

(b)Accept H0

(c)0.0562

Discrete distrbution

20,50,80,100

a,b,c,d

P(S|{X=20})=k/x

P(S∩{X=20}) constant for all x

5(a)Show thtat c=5/8b

(b)2/25,1/5,8/25,2/5

(c)k/100=0.3,k=30,P(S|{X=20})=30/20=1.5>1 so not reasonable

Histogram

N(14.9,3.2²)sth like that, y=kxe⁻ˣ for 0≤x≤4, unit of x is ten years

6(a)0.538

(b)Not symmetrical, median not the same etc

(c)Show that ∫₀ⁿxe⁻ˣdxdx=1-(n+1)e⁻ⁿ

(d)k(∫₀⁴ xe⁻ˣdx)=90 so k=99

(e)0.5902

(f)Sample size=90 is too small to estimate

1D suvat

u=0, t=5, a=3.2

1(a)v=16

(b)s=40

1D forces

5kg, Pulling force=28, a=1.4

2(a)R=49

(b)Friction=21

(c)mu=0.43

Parametric

v=(t²-3t+7)i+(t²-3)j

3(a)|v0|=7.62

(b)t=2

(c)a=(2t-3)i+4tj

(d)t=1.5

Vector suvat

u=-16i-3j, v=2.4,1, t=5, at time T s=4i+cj

4(a)-4i+2j

(b)4.47

(c)T=10,c=10

Projectile

u=28, x=40, y=20

5(a)T=10/(7cosθ)

(b)tan²θ-4tanθ+3=0

(c)36

(d)Spin of the ball, accurate value of g etc

Ladder, tanθ=3/4

6(a)S=1/2Mgcotθ

(b)2/3

(c)√13/3Mg

(d)Farther away so moment increase so S bigger

https://docs.google.com/document/d/12l15wB_3NiOoR2Xb0NrrqihMfVlCsNDnJd-QCWdhw74/edit - here is a link to the UOMS

I wasnt given permission lol

requested at least 3 times

requested at least 3 times

(edited 10 months ago)

Original post by hat25

https://docs.google.com/document/d/12l15wB_3NiOoR2Xb0NrrqihMfVlCsNDnJd-QCWdhw74/edit - here is a link to the UOMS

Anyone can use that?

The person who deleted everything needs to get a life

Quick question, for the normal distribution question was the standard deviation for the hypothesis test root (6.8^2/52) ?

Original post by JustTheFox

Quick question, for the normal distribution question was the standard deviation for the hypothesis test root (6.8^2/52) ?

yes

Original post by Person72

can someone make an UOMS without other people editing it

https://docs.google.com/document/d/1TQGNzDID2rKo_qW3dWAuxMfcLXUhwpU53QmLmVbFZO4/edit

Original post by JustTheFox

Quick question, for the normal distribution question was the standard deviation for the hypothesis test root (6.8^2/52) ?

you had to square root it. What u stated is the variance

Original post by Person72

you had to square root it. What u stated is the variance

Yeah I did? I put root in my message

Original post by JustTheFox

Yeah I did? I put root in my message

my bad, misread it

Last question, for the normal distribution hypothesis question, I just found the critical regions by using the inverse normal option on the calculator and inserting 0.025 and 0.975 as it was two tailed. Is this correct?

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