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Simora [160]
3 years ago
10

The data represents the daily rainfall​ (in inches) for one month. Construct a frequency distribution beginning with a lower cla

ss limit of and use a class width of . Does the frequency distribution appear to be roughly a normal​ distribution?
Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0

Answer:

The frequency distribution does not appear to be normal.

Step-by-step explanation:

The data provided is as follows:

S = {0.38 , 0 , 0.22 , 0.06 , 0 , 0 , 0.21 , 0 , 0.53 , 0.18 , 0 , 0 , 0.02 , 0 , 0 , 0.24 , 0 , 0 , 0.01 , 0 , 0 , 1.28 , 0.24 , 0 , 0.19 , 0.53 , 0 , 0, 0.24 , 0}

It is provided that the first lower class limit should be 0.00 and the class width should be 0.20.

The frequency distribution table is as follows:

Class Interval  Count

 0.00 - 0.19            21

 0.20 - 0.39             6

 0.40 - 0.59             2

 0.60 - 0.79             0

 0.80 - 0.99             0

  1.00 - 1 . 19             0

  1.20 - 1. 39             1

The frequency distribution does not appear to be normal. This is because the frequencies does not start and end at almost equivalent points and the mid-distribution does not consist of the highest frequency.

Thus, the frequency distribution does not appear to be normal.

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3y - 2(60/y) = 6

3y^2-120=6y

3(y^2-40-2y) =0

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6 0
3 years ago
Rafeeq bought a field in the form of a quadrilateral (ABCD)whose sides taken in order are respectively equal to 192m, 576m,228m,
Valentin [98]

Answer:

a. 85974 m²

b. 17,194,800 AED

c. 18,450 AED

Step-by-step explanation:

The sides of the quadrilateral are given as follows;

AB = 192 m

BC = 576 m

CD = 228 m

DA = 480 m

Length of a diagonal AC = 672 m

a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal

The semi-perimeter, s₁,  of ΔABC is found as follows;

s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720

The area, A₁, of ΔABC is given as follows;

Area\, of \, \Delta ABC = \sqrt{s_1\cdot (s_1 - AB)\cdot (s_1-BC)\cdot (s_1 - AC)}

Area\, of \, \Delta ABC = \sqrt{720 \times (720 - 192)\times  (720-576)\times  (720 - 672)}

Area\, of \, \Delta ABC = \sqrt{720 \times 528 \times  144 \times  48} = 6912·√(55) m²

Similarly, area, A₂, of ΔACD is given as follows;

Area\, of \, \Delta ACD= \sqrt{s_2\cdot (s_2 - AC)\cdot (s_2-CD)\cdot (s_2 - DA)}

The semi-perimeter, s₂,  of ΔABC is found as follows;

s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m

We therefore have;

Area\, of \, \Delta ACD = \sqrt{690 \times (690 - 672)\times  (690 -228)\times  (690 - 480)}

Area\, of \, \Delta ACD = \sqrt{690 \times 18\times  462\times  210} = \sqrt{1204988400} = 1260\cdot \sqrt{759} \ m^2

Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square

b. Whereby the cost of 1 meter square land = 200 AED, we have;

Total cost of the land = 200 × 85974 = 17,194,800 AED

c. Whereby the cost of fencing 1 m = 12.50 AED, we have;

Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m

The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED

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