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belka [17]
3 years ago
7

A group surveyed a mix of people below and above 40 years old about whether or not they visit a dentist once a year. This table

gives the survey results.
Visit Dentist Yearly Don’t Visit Dentist Yearly
Below 40 8 22
Above 40 17 13

Which table shows the relative frequency of people above 40 years old who do not visit a dentist once a year? Round your answers to the nearest hundredth.
Mathematics
2 answers:
grigory [225]3 years ago
6 0

Answer with Step-by-step explanation:

              Visit Dentist Yearly Don’t Visit Dentist Yearly

Below 40      8                                        22

Above 40      17                                 13

Relative frequency of people above 40 years old who do not visit a dentist once a year

=Number of people above 40 years old who do not visit a dentist yearly/Number of people above 40

=13/(17+13)

=13/30

=0.43

Hence, Relative frequency of people above 40 years old who do not visit a dentist once a year is:

0.43

il63 [147K]3 years ago
3 0

Answer: is D

                 Visit Dentist Yearly Don’t Visit Dentist Yearly

Below 40                0.27                    0.73

Above 40              0.57                      0.43

Work: 8+22=30        17+13=30     turn into a fraction then simplify to get answer;

22/30   simp= 73.0 (0.73)

13/30     simp= 43.0 (0.43)

8/30      simp= 26.667 (0.27)

17/30      simp= 56.667 (0.57)

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Ostrovityanka [42]

Answer:

No, I'm not old enough to work at taco bell yet

Step-by-step explanation:

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2 years ago
How would the sum of cubes formula be used to factor x3y3 + 343? Explain the process. Do not write the factorization.
Svetllana [295]

Answer:

Given : Expression -x^3y^3+343

To find : How would the sum of cubes formula be used to factor given expression? Explain the process. Do not write the factorization.

Solution :

The formula of sum of cubes is,

a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)

First we convert the given expression in a^3+b^3

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Apply exponent rule : a^mb^m=(ab)^m

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Rewrite 343=7^3

Therefore, x^3y^3+343  could be written as =\left(xy\right)^3+7^3

On comparison with the formula,

a=xy , b=7

\left(xy\right)^3+7^3=(xy+7)(xy^2-xy(7)+7^2)

\left(xy\right)^3+7^3=(xy+7)((xy)^2-7xy+49)            

5 0
3 years ago
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Pani-rosa [81]

Answer:

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Step-by-step explanation:

Answer:

-6 3/8 < -7/8

Showing Work

Using the given inputs:

-6 3/8 -7/8

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Answer:

m=\dfrac{3}{2}

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}

<u>1) Finding (\overline{x},\overline{y})</u>

  • Average of the x coordinates:

\overline{x} = \dfrac{1+2+3}{3}

\overline{x} = 2

  • Average of the y coordinates:

similarly for y

\overline{y} = \dfrac{3+3+6}{3}

\overline{y} = 4

<u>2) Finding the line through (\overline{x},\overline{y}) with slope m.</u>

Given a point and a slope, the equation of a line can be found using:

(y-y_1)=m(x-x_1)

in our case this will be

(y-\overline{y})=m(x-\overline{x})

(y-4)=m(x-2)

y=mx-2m+4

this is our equation of the line!

<u>3) Find the squared vertical distances between this line and the three points.</u>

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.  

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We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

y=m(1)-2m+4

y=-m+4

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

d_1=3-(-m+4)

d_1=m-1

finally, as asked, we'll square the distance

(d_1)^2=(m-1)^2

  • Distance from point (2,3)

we'll do the same as above here:

y=m(2)-2m+4

y=4

vertical distance between the two points: (2,3) and (2,4)

d_2=3-4

d_2=-1

squaring:

(d_2)^2=1

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y=m(3)-2m+4

y=m+4

vertical distance between the two points: (3,6) and (3,m+4)

d_3=6-(m+4)

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3) Add up all the squared distances, we'll call this value R.

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Looking at the equation above, we can tell that R is a function of m:

R(m)=(m-1)^2+4+(2-m)^2

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)

\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)

now to find the minimum value we'll just use a condition that \dfrac{dR}{dm}=0

0=2(m-1)+2(2-m)(-1)

now solve for m:

0=2m-2-4+2m

m=\dfrac{3}{2}

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

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Answer:

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Distribute the equation so that B and C are by itself.

New Equation: AxB + AxC

This is distributive property.

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