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Volgvan
3 years ago
8

The probability of a law school graduate passing the Bar exam is 0.43. In a graduating

Mathematics
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

hbhbhbb dsvjv wenvwe

Step-by-step explanation:

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Ill give 50 points ik i just asked but i rlly need help
belka [17]

Answer:

2 runners will be needed

Step-by-step explanation:

1/2 = 2/4

2/4 / 1/4 = 2

2 runners, for 1/2 mile

8 0
3 years ago
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If = and x > 0 what is the value of x ?
Masteriza [31]

O valor de x é 45º

Explicação:Confia no pai

8 0
3 years ago
Miguel run faster than 19 of his teammates what percent of the team is slower than Miguel
nika2105 [10]

Answer:

he is faster than 95% of his team

Step-by-step explanation:

To finde this

First: Add Miguel to his 19 people equals 20 people

Next divide the 19/20

Providing the answer 0.95

to get the percentage multiply the decimal by 100: 0.95*100

= 95%

4 0
3 years ago
Write a description of the rule (x, y) = (x - 4, y + 3).
storchak [24]

9514 1404 393

Answer:

  translate 4 left and 3 up

Step-by-step explanation:

The value added to the x-coordinate is the translation to the right. Here, it is -4, so the translation is 4 units left.

The value added to the y-coordinate is the translation up. Here, it is +3, so the translation is 3 units up.

The rule effects a translation 4 units left and 3 units up.

4 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

4 0
3 years ago
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