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olga2289 [7]
3 years ago
13

Nidhi is creating a rectangular garden in her backyard. The length of the garden is 10 feet. The perimeter of the garden must be

at least 40 feet and no more than 76 feet. Use a compound inequality to find the range of values for the width w of the garden.
Fill in the blanks
__ ≤ w ≤ __
Mathematics
1 answer:
Basile [38]3 years ago
5 0

Answer:

The range of width is 10 ≤ W ≤ 28

Step-by-step explanation:

* lets study the meaning of compound inequality

- If x is greater than a and x is smaller than b, then x is between a and b

∵ x > a and x < b

∴ The compound inequality is ⇒ a < x < b

# Ex: ∵ x > -2 and x < 10

∴ The compound inequality is ⇒ -2 < x < 10

- If x is greater than or equal a and x is smaller than or equal b, then

 x is from a and b

∵ x ≥ a and x ≤ b

∴ The compound inequality is ⇒ a ≤ x ≤ b

# Ex: ∵ x ≥ -2 and x ≤ 10

∴ The compound inequality is ⇒ -2 ≤ x ≤ 10

* Now lets solve the problem

- The garden in the shape of a rectangle with dimensions length (L)

 and width (W)

- The length of the garden is 10 feet

- The perimeter (P) of the garden is at least 40 feet and not more than

  76 feet

∵ L = 10 feet

∵ P = 2L + 2W

- At least means greater than or equal (≥) and not more than means

 smaller than or equal (≤)

∴ P ≥ 40 feet

∴ P ≤ 76 feet

- lets use the rule of the perimeter

∴ 2(10) + 2(W) ≥ 40 ⇒ simplify

∴ 20 + 2W ≥ 40 ⇒ subtract 20 from both sides

∴ 2W ≥ 20 ⇒ divide both sides by 2

∴ W ≥ 10 ⇒ (1)

- Do similar with P ≤ 76

∴ 2(10) + 2(W) ≤ 76 ⇒ simplify

∴ 20 + 2W ≤ 76 ⇒ subtract 20 from both sides

∴ 2W ≤ 56 ⇒ divide both sides by 2

∴ W ≤ 28 ⇒ (2)

- From (1) and (2)

∴ 10 ≤ W ≤ 28 ⇒ compound inequality

* The range of the width is from 10 feet to 28 feet

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Given that lie <span>detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

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</span>P(X)={ ^nC_xp^xq^{n-x}} \\  \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\  \\ =1\times0.0874\times1=0.0874
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</span>Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
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P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\  \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\  \\ =0.9126


Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

\mu=npq \\  \\ =15\times0.15\times0.85 \\  \\ =1.9125


Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

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