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OleMash [197]
3 years ago
7

Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an ove

rlayed frequency polygon (A.K.A.-line graph) and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?
Life Expectancy at Birth – Women Frequency
49–55 1
56–62 2
63–69 1
70–76 3
77–83 9
84–90 4
Table 2.47
Life Expectancy at Birth – Men Frequency
49–55 3
56–62 3
63–69 2
70–76 3
77–83 5
84–90 4
Mathematics
1 answer:
sweet [91]3 years ago
3 0

Answer:

<em>conclusion : Life expectancy of women > Life expectancy of men </em>

Step-by-step explanation:

A) Mean life expectancy of women

from the attached table

mean life expectancy of women = ∑fx / ∑f

∑fx = 1523

∑f = 20

mean = 1523 / 20 = 76.15

The line graph shows a negatively skewed distribution

B) Mean life expectancy of Men

from the attached table

mean life expectancy of Men = ∑fx / ∑f

∑fx = 1432

∑f = 20

mean = 1432 / 20 = 71.6

<em>conclusion : Life expectancy of women > Life expectancy of men </em>

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Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

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Help ASAP!!!<br> First to answer with a correct answer gets brain...
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Hey there!

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A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function
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Here's our equation.

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To find this out, we can plug in 0 and solve for t.

0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a}  \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16}  = \frac{64\±\sqrt{4096+192}}{32}

= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}

So the ball will return to the ground at the positive value of \boxed{\frac{8+\sqrt{67}}{4}} seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2

h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}


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