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Bond [772]
3 years ago
15

The boxplots below display annual incomes (in thousands of dollars) of households in two cities. Which city has a greater percen

tage of households with annual incomes above $80,000?
a. Statstown
b. Medianville
c. Both cities have about the same percentage. of households with annual incomes above $80,000.
d. It is impossible to tell from the boxplots.

Mathematics
1 answer:
kompoz [17]3 years ago
4 0

Answer:

Option C is correct.

Both cities have about the same percentage. of households with annual incomes above $80,000.

Step-by-step explanation:

The boxplot to read answers from, seem to be missing. On checking the question online, the boxplot for the question was obtained.

This boxplot is attached to the solution of this question.

On the boxplots, it is evident that the two cities have an average of $80000 because the line in the middle of boxplots point out the median. And boxplots only show the spread of data over a range, since the median for the two cities are both $80000, the percentage of households with income greater than $80000 in both cities is the same and that percentage is roughly 50% in each of the two cities.

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Every week, the mass of the sample is multiplied by a factor of 0.81

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2 years ago
What are the solutions to the equation
frosja888 [35]

Answer:

C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

Step-by-step explanation:

You have the quadratic function 2x^2-x+1=0 to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions ax^2+bx+c=0 with a\neq 0 the Bhaskara's Formula is:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}

x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}

It usually has two solutions.

Then we have  2x^2-x+1=0  where a=2, b=-1 and c=1. Applying the formula:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}

Observation: \sqrt{-1}=i

x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i

And,

x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i

Then the correct answer is option C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

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3 years ago
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