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3 years ago

An engineering firm is hired to determine if certain waterways in Virginia are safe for fishing. Samples

Mathematics

1 answer:

Natalija [7]3 years ago

6 0

Answer:

(a) S = {FFF, FFN, FNN, FNF, NNN, NNF, NFF, NFN}

(b) S(E) = {FFF, FFN, FNF, NFF}

Step-by-step explanation:

F = safe to fish

N = not safe to fish

number of rivers sampled = 3

(a) All of the possible outcomes obtained from the samples are listed in the sample space bellow:

S = {FFF, FFN, FNN, FNF, NNN, NNF, NFF, NFN}

(b) If at least two rivers are safe, the sample space for E is all of the events with two or three safe rivers:

S(E) = {FFF, FFN, FNF, NFF}

(c) In the set {FFF, NFF, FFN, NFN}, the second lake sample is safe to fish in all of the elements. The event is: second river is safe for fishing.

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Answer:

the first one

Step-by-step explanation:

the others don't make any sense and also the first one's the only one that's in inequality form.

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3 years ago

Read 2 more answers

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

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3 years ago

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Answer:

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Step-by-step explanation:

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