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Vesna [10]
3 years ago
13

Malcolm is trying a very low-carbohydrate diet. He would like to keep the amount of carbs consumed in grams between the levels s

hown in the following compound inequality:
50 < 2x + 10 and 2x + 10 < 110

Solve for x in this inequality, and explain what the answer represents.

1. x < 20 and x > 50; Malcolm needs to consume less than 20 grams of carbohydrates or more than 50 grams of carbohydrates.
2. x > 20 and x < 50; Malcolm needs to consume more than 20 grams of carbohydrates, but less than 50 grams of carbohydrates.
3. x > 30 and x < 60; Malcolm needs to consume more than 30 grams of carbohydrates, but less than 60 grams of carbohydrates.
3. x < 30 and x > 60; Malcolm needs to consume less than 30 grams of carbohydrates or more than 60 grams of carbohydrates.
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Answer:

The correct option is 2.

Step-by-step explanation:

The given inequalities are

50             .... (1)

2x+10            .... (2)

Solve each inequality.

Subtract 10 from each side of inequality (1).

50-10

40

Divide both the sides by 2.

20

The value of x is more than 20.

Subtract 10 from each side of inequality (2).

2x+10-10

2x

Divide both the sides by 2.

x

The value of x is less than 50.

Since x > 20 and x < 50, therefore Malcolm needs to consume more than 20 grams of carbohydrates, but less than 50 grams of carbohydrates.

Hence the correct option is 2.

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Answer:

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Step-by-step explanation:

Given the integrand \int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx, before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.

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Integrating its partial sum

\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx  = \int\limits  (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2}  \int\limits{\frac{1}{x} } \, dx  + \frac{2}{13}  \int\limits{\frac{1}{x-2} } \, dx

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5 0
3 years ago
Airplanes approaching the runway for landing are required to stay within the localizer (a certain distance left and right of the
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Answer:

a) P(x = 0) = 64.69%

b) P(x ≥ 1) = 35.31%

c) E(x) = 0.42

d) var(x) = 0.3906

Step-by-step explanation:

The given problem can be solved using binomial distribution since:

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The binomial distribution is given by

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ

Where n is the number of trials, x is the variable of interest and p is the probability of success.

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Here we have x = 0, n = 6 and p = 0.07

P(x = 0) = ⁶C₀(0.07⁰)(1 - 0.07)⁶⁻⁰

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The probability that at least 1 plane exceeds the localizer is given by

P(x ≥ 1) = 1 - P(x < 1)

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Therefore, the expected number of planes to exceed the localizer on any given day is 0.42

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The variance for the number of planes to exceed the localizer is given by

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