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Aloiza [94]
2 years ago
12

Help Me please . I want to know Square root of 169 and 144 and 625 . please man fast ​

Mathematics
2 answers:
irga5000 [103]2 years ago
6 0

Answer:

\sqrt{169} = 13\\\\\sqrt{144} = 12\\\\\sqrt{625} = 25

Step-by-step explanation:

\sqrt{169} = \left(13^2 \right)^{\tfrac 12} = 13^{\tfrac 22} = 13^1 = 13\\\\\sqrt{144} = \left( 12^2\right)^{\tfrac 12} = 12^{\tfrac 22} = 12^1 = 12\\\\\sqrt{625}= \left(25^2 \right)^{\tfrac 12} = 25^{\tfrac 22} = 25^1 = 25

neonofarm [45]2 years ago
5 0

Answer:

13,12,25

Step-by-step explanation:

Square root of 169 is 13.

Square root of 144 is 12

Square root of 625 is 25.

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I think this is right:

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During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for
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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} } }]

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    = [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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