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gizmo_the_mogwai [7]
1 year ago
5

​(c) if the corresponding terms of two arithmetic sequences are​ added, is the resulting sequence​ arithmetic?

Mathematics
1 answer:
Harman [31]1 year ago
5 0

To determine whether the corresponding terms of 2 arithmetic sequence's added will give new arithmetic sequence or not, Let' take 2 Arithmetic sequences.

In one first term is a1 and common difference is d1, in the other first term is a2 and common difference is d2.

Now nth term for first sequence = a1+(n-1) d1

  nth term for second sequence = a2+(n-1) d2

Now add the 2 terms: a1+(n-1)d1 +a2 +(n-1)d2

                                   = a1+a2 + (n-1)(d1+d2)

This is again new arithmetic sequence with first term a1+a2 and common difference d1+d2.

Hence if we add corresponding terms of 2 arithmetic sequence, we will again get an arithmetic sequence.

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Step-by-step explanation:

It is given that triangle MRN is created when an equilateral triangle is folded in half.

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2 years ago
A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta
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Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true proportion 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 of women who find success with this new treatment = \frac{7}{14} = 0.50

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

So, 95% confidence interval for the true 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 />

<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} } }]

= [ 0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } , 0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } ]

 = [0.238 , 0.762]

Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

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