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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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Answer:

-51t+45

Step-by-step explanation:

-6(3t-6)-3(11t-3)

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6 0
3 years ago
Substitute 9 into each equation to determine if 9 is the solution. Drag each equation into the correct column.
IRISSAK [1]

Answer:

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Those can all be solved with 9

the other can not

Step-by-step explanation:

i hope this helped

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3 years ago
Mary has two job offers.
galina1969 [7]

Answer:

8h+100

8.50h+80

Step-by-step explanation:

7 0
3 years ago
Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

3 0
3 years ago
Solve this please<br><img src="https://tex.z-dn.net/?f=4m%20-%205%20%3D%2011" id="TexFormula1" title="4m - 5 = 11" alt="4m - 5 =
krek1111 [17]
M has to equal 2. 11-5 equals 6. 6-4 is 2. 2 is the missing value.
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