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dimulka [17.4K]
3 years ago
7

x + 8(x + 2) = 53

Mathematics
1 answer:
pishuonlain [190]3 years ago
5 0
X+8(x+2)=53
 2x+10=53  
     - 10| -10
------------------
       2x= 43
        /2    /2
        x = 21.5  
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A certain type of bird has a probability of 1/12 Harding a bird with white feather. If a bird lays 2 eggs, find the probability
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A. \frac{1}{12}*(1-\frac{1}{12})\\\frac{1}{12}*\frac{11}{12}\\\frac{11}{144}

B. \frac{1}{12}*(1-\frac{1}{12})+\frac{1}{12}*\frac{1}{12}\\\frac{1}{12}*\frac{11}{12}+\frac{1}{144}\\\frac{11}{144}+\frac{1}{144}\\\frac{12}{144}\\\frac{1}{12}
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Single adults: According to a Pew Research Center analysis of census data, in 2012, 20% of American adults ages 25 and older had
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Option B

Step-by-step explanation:

The number that had never been married will vary in each sample due to the random selection of adults.

This number will vary in each sample to the random selection process but they might or might not be as close as possible to one another after sampling.

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A pizza company uses equation C = 15n to calculate the cost of buying a number of pizzas. What is the constant of proportionalit
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29

Step-by-step explanation:

7 0
3 years ago
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Find the equation of the line parallel to y = -3x + 4 with a y-intercept of -10
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8 0
3 years ago
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An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = \sqrt{(15-x)^2+1^2}

The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

6 0
3 years ago
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