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lozanna [386]
3 years ago
14

2/3 + 1/2 i just want to give some points away

Mathematics
2 answers:
goldfiish [28.3K]3 years ago
6 0
Answer: 7/6, 1 and 1/6, or 1.16


Step by step explanation:

2/3 + 1/2

4+3/6

7/6
KatRina [158]3 years ago
4 0

Answer:

7/6 or 1 and 1/6

Hope this helps! <33

Lol I'm answering this so late-

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Add −3x2y2z2+12xyz2+7xyz and 10xyz−10xyz2
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Answer:

3xyz

Step-by-step explanation:

The required sum = 7xyz+(−5xyz)+9xyz+(−8xyz)

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=(7–5+9–8)xyz

=(16–13)xyz

=3xyz

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Combine like terms <br> 9j+34j
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During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for
AlexFokin [52]

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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3 years ago
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Dmitrij [34]

The statement is true. X= all real numbers

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