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Harlamova29_29 [7]
3 years ago
5

Which table of ordered pairs, when plotted, will form a straight line? Select two answers.

Mathematics
1 answer:
aniked [119]3 years ago
4 0

Answer:

Options A and D are answers.

Step-by-step explanation:

The ordered pairs are given in the table in options. We have to choose which table gives a straight line.

We know that if the increase or decrease of y value with respect to corresponding x values is uniform then the plot of those x and y corresponding values will form a straight line.

Now, in table A, the slope of each pair of x and y values is constant i.e.  

\frac{- 1 - 0}{- 2 - ( - 1)} = \frac{0 - 1}{- 1 - 0} = \frac{1 - 6}{0 - 5} = 1

Hence, it will give a straight line.

In table B, the slope of each pair of x and y values is not constant i.e.

\frac{7 - (- 3)}{- 2 - 0} \neq  \frac{- 3 - 12}{0 - 3}.

Hence, the plotted graph will not give a straight line.

In table C, the slope of each pair of x and y values is not constant i.e.

\frac{- 3 - 5}{- 4 - 0} = \frac{5 - 9}{0 - 2} \neq  \frac{9 - 12}{2 - 5}

Hence, it will not give a straight line.

In table D,  the slope of each pair of x and y values is constant i.e.

\frac{-8 - (- 4)}{- 2 - 0} = \frac{- 4 - 4}{0 - 4} = \frac{4 - 8}{4 - 6}

Hence, it will give a straight line.

So, options A and D answer.

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Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is :  \hat p = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

      P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion = 0.345

           n = sample of participants = 3532

           p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population​ proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                         significance are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u>= [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

    = [ 0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } , 0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } ]

    = [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

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