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Anna007 [38]
4 years ago
14

Laura has moved to a new apartment. Her schoolbooks comprising of different subjects are mixed in a bag during the move. Four bo

oks are of mathematics, three are English, and six are science. If Laura opens the bag and selects books at random, find the given probability.
P(2 science and 2 mathematics books)
Mathematics
2 answers:
Inessa05 [86]4 years ago
8 0
MATH: 4 out of 13
ELA : 3 out of 13
SCIENCE: 6out of 13
monitta4 years ago
6 0
6/13 * 5/12 * 4/11 * 3/10 = <span>0.02097902097 = 2.1%</span>
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Alright, second try. 15 Points to whoever can solve with a half-decent explanation.
sesenic [268]

Answer:

29 cans

Step-by-step explanation:

To solve this problem we must use Heron's formula; A = \sqrt{S(S-a)(S-b)(S-c)}

S = 1/2 the perimeter of the given triangle

a, b, c = the measurements of the three given sides

The first step to solving this question is first finding the perimeter.

P = 65m + 74m + 67m

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Now that we have our perimeter, we must divide P by 2 to get S.

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A = \sqrt{103(103-65)(103-67)(103-74)}

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A = \sqrt{4086216}

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A ≈ 2021.43909134... m²

If a can of paint covers 70 square meters, then you will need to figure out how many cans you need. This can be done by dividing the overall area of the triangle by the area that a single can of paint can cover.

\frac{2021.43909134...}{70} = 28.8777013049...

We know we can't buy 28.87... of a can (people simply do not sell parts of cans or half-full containers... unless its a bag of lays chips and their 80% air nonsense...) so we must round up to the nearest whole number.

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Step-by-step explanation:

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expeople1 [14]

Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

AD = 15, AB = 12, BC = 15, CD = 12

By Pythagorean Theorem, we determine the length of the line segment BD:

BD = \sqrt{AB^{2}+AD^{2}} (1)

BD = \sqrt{12^{2}+15^{2}}

BD = 3\sqrt{41}

In addition, we know the following characteristics of the rectangle BEFD:

BD = EF, EF = EC + CF, BE = FD (2), (3), (4)

By Pythagorean Theorem:

BC^{2} = BE^{2}+EC^{2} (5)

CD^{2} = CF^{2}+DF^{2} (6)

By (3), (4), (5) and (6):

BC^{2} = BE^{2} + EC^{2} (7)

CD^{2} = (EF-EC)^{2} + BE^{2} (8)

By (7) in (8):

CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})

CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}

CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}

Then, we clear EC:

2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}

EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}

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EC = \frac{75\sqrt{41}}{41}

And the length of the line segment CF is:

CF = EF - EC

CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}

CF = \frac{48\sqrt{41}}{41}

And the length of the line segment DF is determined by Pythagorean Theorem:

FD = \sqrt{CD^{2}-CF^{2}}

FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}

FD = \frac{60\sqrt{41}}{41}

And the area of the rectangle is determined by the following formula:

A = FD\cdot EF

A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})

A = 180

The area of rectangle BEFD is 180 square units.

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3 years ago
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