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NECO GCE 2017 General Mathematics Answer Now Available

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1-10 = CEDADDAEDB

11-20= DCBEBCACEC

21-30= CBD_BEBBAD

31-40= DC_BDAD_A_

41-50= D_DACCACC_

51-60= BAAADCD_BA

————————–

5b)Mode = L1 + (i(f1 – f0) / (2f1 – f0 – f2))

Where:

L1 = Lower limit of the modal class

f1 = Frequency of the modal class

f0 = Frequency of the class preceding the modal class

f2 = Frequency of the class succeeding the modal class

i = Class interval

i = 10

L1 = 30.5

f1 = 15

f0 = 12

f2 = 7

Mode = L1 + (i(f1 – f0) / (2f1 – f0 – f2))

Mode = 30.5 + (10(15 – 12) / (2(15) – 12 – 7))

Mode = 30.5 + (10(3) / (30 – 12 – 7))

Mode = 30.5 + (30 / 11)

Mode = 30.5 + 2.73

Mode = 33.23

5ci)Mean = ∑fm / ∑f

∑fm = 33 + 155 + 306 + 532.5 + 301

∑fm = 1327.5

∑f = 6 + 10 + 12 + 15 + 7

∑f = 50

Mean = ∑fm / ∑f

Mean = 1327.5 / 50

Mean = 26.55

5cii5b)5cii)5c)L ₁  + i(((N / 2) – cf) / f)

Where:

L ₁  = Lower limit of the median class

f = Frequency of the median class

cf = Cumulative frequency of the class preceding the median class

i = Class interval of the median class

N = Sum of the Frequencies = 49

Data table
x f cf C.B
1-10 6 6 0.5 -10.5

11-20 10 16 10.5 – 20.5

21-30 11 27 20.5 – 30.5

31-40 15 42 30.5 – 40.5

41-50 7 49 40.5 – 50.5
5c)L ₁  + i(((N / 2) – cf) / f)

Where:

L ₁  = Lower limit of the median class

f = Frequency of the median class

cf = Cumulative frequency of the class preceding the median class

i = Class interval of the median class

N = Sum of the Frequencies = 49

Data table
x f cf C.B
1-10 6 6 0.5 -10.5

11-20 10 16 10.5 – 20.5

21-30 12 28 20.5 – 30.5

31-40 15 43 30.5 – 40.5

41-45 7 50 40.5 – 50.5

Median Class = 20.5 – 30.5

L1 = 20.5

N = 50

cf = 16

i = 10

f = 12

Median = L1 + i(((N / 2) – cf) / f)

Median = 20.5 + 10(((50 / 2) – 16) / 12)

Median = 20.5 + 10((25 – 16) / 12)

Median = 20.5 + 10(9 / 12)

Median = 20.5 + 10(0.75)

Median = 20.5 + 7.5

Median = 28

12c)Standard Deviation = √(∑f|m – Mean|² / N)

Where:

m = Mid value of each grouped data

N = Sum of the frequencies

Mean = ∑fm / N

N = 12 + 35 + 21 + 22 + 10

N = 100

∑fm = 66 + 542.5 + 535.5 + 781 + 455

∑fm = 2380

Mean = 2380 / 100

Mean = 23.8

∑f|m – Mean|² = 4018.68 + 2411.15 + 60.69 + 3011.58 + 4708.9

∑f|m – Mean|² = 14211

Standard Deviation = √(∑f|m – Mean|² / N)

Standard Deviation = √(14211 / 100)

Standard Deviation = √(142.11)

Standard Deviation = 11.92

11b) DISTANCE WE ALONG THE LINE OF LATITUDE

θ = 42° + 22°

θ = 64°

Therefore, your angular difference = 64°
QR = (θ⁄360) x 2πR

Where:

y = Distance Along Two Great Circles

θ = Angular Difference

R = Radius of the Earth

QR= (θ⁄360) x 2πR

QR = (64⁄360) x (2 x π x 6400)

QR = 0.178 x 40212.39

QR = 7148.87Km

11a) DISTANCE PW ON A PARRALLEL OF LATITUDE

P(42°N ,48°E) and Q(42°N, 36°W)

θ = 48° + 36°

θ = 84°

operation is y = (θ⁄360) x 2πR

Where:

y = Distance Along Two Great Circles

θ = Angular Difference

R = Radius of the Earth

PQ= (θ⁄360) x 2πR

pQ= (84⁄360) x (2 x π x 6400)

PQ = 0.233 x 40212.39

PQ= 9382.89Km.

12a) Mean = ∑fm / ∑f

x f. m. Fm
1-10 12 5.5 66

11-20 35 15.5 542.5

21-30 21 25.5 535.5

31-40 22 35.5 781

41-50 10 45.5 455

∑fm = 66 + 542.5 + 535.5 + 781 + 455

∑fm = 2380

∑f = 12 + 35 + 21 + 22 + 10

∑f = 100

Mean = ∑fm / ∑f

Mean = 2380 / 100

Mean = 23.8

2b) x + y =3______(1)
x² – y² = 15_____(2)

Solution

From equation (2)

x² – y² = 15

(x+y)(x-y) = 15_____(3)

Recall x-y = 3

Substitute the value of (x-y) in equation (3)

3(x + y) = 15

Therefore (x +y) =5

The value of x+y =5

12b)Mean Deviation = ∑f|m – Mean| / N

Where:

m = Mid value of the grouped data

N = Sum of the frequencies

Mean = ∑fm / N

N = 12 + 35 + 21 + 22 + 10

N = 100

∑fm = 66 + 542.5 + 535.5 + 781 + 455

∑fm = 2380

Mean = 2380 / 100

Mean = 23.8

Mean Deviation = ∑f|m – Mean| / N

∑f|m – Mean| = 219.60 + 290.5 + 35.69+ 257.4 + 217

∑f|m – Mean| = 1020.19

Mean Deviation = 1020.19 / 100

Mean Deviation = 10.20

4b)A = πrl + πr²

Where:

A = Area of the Cone
r = Radius of the Cone
l = Slant Height of the Cone

A = (π x 7 x 7) + (π x 7²)

A = 153.94 + (π x 49)

A = 153.94 + 153.94

A = 307.88cm²

7b)
y²/36 – 1/9 = 0

Multiply through by 36

y² – 4 = 0

y² = 4

y =±2
[11/11, 10:54] ‪+234 703 944 4757‬: 9a)Log₄(x² + 7x + 28) = 2

Solution

Log₄(x² + 7x + 28) = 2

x² + 7x + 28 = 2⁴

x² + 7x + 28 = 16

x² + 7x + 12 =0

((-b ± √(b² – 4ac)) / 2a)

where:

a = coefficient of x²
b = coefficient of x and
c = constant

((-7 ± √(7² – 4 x 1 x 12)) / (2 x 1))

((-7 ± √(49 – 48)) / 2)

((-7 ± √(1)) / 2)

(-7 ± 1) / 2

(-7 + 1) / 2 or (-7 – 1) / 2

(-6) / 2 or (-8) / 2

-3 or -4

9a)Log₄(x² + 7x + 28) = 2

Solution

Log₄(x² + 7x + 28) = 2

x² + 7x + 28 = 2⁴

x² + 7x + 28 = 16

x² + 7x + 12 =0

((-b ± √(b² – 4ac)) / 2a)

where:

a = coefficient of x²
b = coefficient of x and
c = constant

((-7 ± √(7² – 4 x 1 x 12)) / (2 x 1))

((-7 ± √(49 – 48)) / 2)

((-7 ± √(1)) / 2)

(-7 ± 1) / 2

(-7 + 1) / 2 or (-7 – 1) / 2

(-6) / 2 or (-8) / 2

-3 or -4

4a) length of a chord =2rSinΘ/2

where

Θ=94°
circle of radius =4cm

= 2(4) Sin94/2

=8Sin(47)

= 8(0.7314)

=5.85cm

n(µ)=40 = sum of subset

40 = 35 -x +x +26-x

40 = 61 -x

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Posted by Mizter AGUNS On November 11, 2017

Categories: neco

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