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il63 [147K]
2 years ago
6

Rewrite 32/35 and 9/10

Mathematics
1 answer:
hram777 [196]2 years ago
6 0

The question was incorrect. Please find the correct content below.

Compare 32/35 and 9/10.

Comparing 32/35 and 9/10 we have 32/35 is greater than 9/10, that is 32/35 > 9/10.

Fraction is the ratio of two numbers. The upper number is called Numerator and the Lower number is called the Denominator.

We know that if the denominators are the same for two fractions then which has the greatest numerator is a greater fraction than the other.

Given the fractions are 32/35, 9/10

To compare this two fractions we have to make denominators equal first.

LCM of 10,35 = 70

Calculating the fractions,

32/35 = (32*2)/(35*2) = 64/70

9/10 = (9*7)/(10*7) = 63/70

Since 64 > 63

So 64/70 > 63/70

Therefore, 32/35 > 9/10

Hence fraction 32/35 is greater than the other fraction 9/10.

Learn more about Fraction here -

brainly.com/question/78672

#SPJ10

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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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