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3 years ago

X f(x) -2 -1/200 -1 -1/20 0 -1/2 1 -5 2 -50

Mathematics

1 answer:

Hitman42 [59]3 years ago

6 0

9514 1404 393

Answer:

f(x) = -1/2·10^x

Step-by-step explanation:

The y-intercept is -1/2.

The common ratio is (-1/20)/(-1/200) = 10, so the exponential function can be written as ...

f(x) = -1/2·10^x

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In the past decades there have been intensive antismoking campaigns sponsored by both federal and private agencies. In one study

DIA [1.3K]

Answer:

$z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055$

The p value can be calculated from the alternative hypothesis with this probability:

$p_v =2*P(Z>2.055)=0.0399$

And the best option for this case would be:

C. between 0.01 and 0.05.

Step-by-step explanation:

Information provided

$X_{1}=1642$ represent the number of smokers from the sample in 1995

$X_{2}=1415$ represent the number of smokers from the sample in 2010

$n_{1}=4276$ sample from 1995

$n_{2}=3908$ sample from 2010

$p_{1}=\frac{1642}{4276}=0.384$ represent the proportion of smokers from the sample in 1995

$p_{2}=\frac{1415}{3908}=0.362$ represent the proportion of smokers from the sample in 2010

$\hat p$ represent the pooled estimate of p

z would represent the statistic

$p_v$ represent the value for the pvalue

System of hypothesis

We want to test the equality of the proportion of smokers and the system of hypothesis are:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

The statistic is given by:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{1642+1415}{4276+3908}=0.374$

Replacing the info given we got:

$z=\frac{0.384-0.362}{\sqrt{0.374(1-0.374)(\frac{1}{4276}+\frac{1}{3908})}}=2.055$

The p value can be calculated from the alternative hypothesis with this probability:

$p_v =2*P(Z>2.055)=0.0399$

And the best option for this case would be:

C. between 0.01 and 0.05.

7 0

3 years ago

Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).

ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

$\overrightarrow {PQ}==$

$\overrightarrow {PR}==$

The nonzero vector orthogonal to the plane through the points P,Q, and R is

$\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}$

$\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}$

Expand along row 1.

$\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}$

$\overrightarrow n=i(-9)-j(8)+k(2)$

$\overrightarrow n=-9i-8j+2k$

$\overrightarrow n=$

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

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3 years ago

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Oksanka [162]

Answer:

x = 12

(getting to the 20 character limit, please ignore this)

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The correct answer is D

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Oxana [17]

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Step-by-step explanation:

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